Vector Calculus
Fourth Edition
if* Jr -p
Susan Jane Colley
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Vector Calculus
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Vector Calculus! 4™
Susan Jane Colley
Oberlin College
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Library of Congress Cataloging-in-Publication Data
Colley, Susan Jane.
Vector calculus / Susan Jane Colley. - 4th ed.
p. cm.
Includes index.
ISBN-13: 978-0-321-78065-2
ISBN-10: 0-321-78065-5
1. Vector analysis. I. Title.
QA433.C635 2012
515'.63-dc23
2011022433
Copyright © 2012, 2006, 2002 Pearson Education, Inc.
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ISBN 13: 978-0-321-78065-2
ISBN 10: 0-321-78065-5
To Will and Diane,
with love
About the Author
Susan Jane Colley
Susan Colley is the Andrew and Pauline Delaney Professor of Mathematics at
Oberlin College and currently Chair of the Department, having also previously
served as Chair.
She received S.B. and Ph.D. degrees in mathematics from the Massachusetts
Institute of Technology prior to joining the faculty at Oberlin in 1983.
Her research focuses on enumerative problems in algebraic geometry, partic-
ularly concerning multiple-point singularities and higher-order contact of plane
curves.
Professor Colley has published papers on algebraic geometry and commuta-
tive algebra, as well as articles on other mathematical subjects. She has lectured
internationally on her research and has taught a wide range of subjects in under-
graduate mathematics.
Professor Colley is a member of several professional and honorary societies,
including the American Mathematical Society, the Mathematical Association of
America, Phi Beta Kappa, and Sigma Xi.
Contents
Preface ix
To the Student: Some Preliminary Notation xv
1 Vectors 1
1.1 Vectors in Two and Three Dimensions 1
1.2 More About Vectors 8
1.3 The Dot Product 18
1.4 The Cross Product 27
1.5 Equations for Planes; Distance Problems 40
1.6 Some n-dimensional Geometry 48
1.7 New Coordinate Systems 62
True/False Exercises for Chapter 1 75
Miscellaneous Exercises for Chapter 1 75
Differentiation in Several Variables 82
2.1 Functions of Several Variables; Graphing Surfaces 82
2.2 Limits 97
2.3 The Derivative 116
2.4 Properties; Higher-order Partial Derivatives 134
2.5 The Chain Rule 142
2.6 Directional Derivatives and the Gradient 158
2.7 Newton's Method (optional) 176
True/False Exercises for Chapter 2 182
Miscellaneous Exercises for Chapter 2 183
3 Vector-Valued Functions 189
3.1 Parametrized Curves and Kepler's Laws 189
3.2 Arclength and Differential Geometry 202
3.3 Vector Fields: An Introduction 221
3.4 Gradient, Divergence, Curl, and the Del Operator 227
True/False Exercises for Chapter 3 237
Miscellaneous Exercises for Chapter 3 237
4 Maxima and Minima in Several Variables 244
4.1 Differentials and Taylor's Theorem 244
4.2 Extrema of Functions 263
4.3 Lagrange Multipliers 278
4.4 Some Applications of Extrema 293
True/False Exercises for Chapter 4 305
Miscellaneous Exercises for Chapter 4 306
Multiple Integration 310
5.1 Introduction: Areas and Volumes 310
5.2 Double Integrals 314
5.3 Changing the Order of Integration 334
5.4 Triple Integrals 337
5.5 Change of Variables 349
5.6 Applications of Integration 373
5.7 Numerical Approximations of Multiple Integrals (optional) 388
True/False Exercises for Chapter 5 401
Miscellaneous Exercises for Chapter 5 403
6 Line Integrals 408
6.1 Scalar and Vector Line Integrals 408
6.2 Green's Theorem 429
6.3 Conservative Vector Fields 439
True/False Exercises for Chapter 6 450
Miscellaneous Exercises for Chapter 6 451
7 Surface Integrals and Vector Analysis 455
7.1 Parametrized Surfaces 455
7.2 Surface Integrals 469
7.3 Stokes's and Gauss's Theorems 490
7.4 Further Vector Analysis; Maxwell's Equations 510
True/False Exercises for Chapter 7 522
Miscellaneous Exercises for Chapter 7 523
8 Vector Analysis in Higher Dimensions 530
8.1 An Introduction to Differential Forms 530
8.2 Manifolds and Integrals of k-forms 536
8.3 The Generalized Stokes's Theorem 553
True/False Exercises for Chapter 8 561
Miscellaneous Exercises for Chapter 8 561
Suggestions for Further Reading 563
Answers to Selected Exercises 565
Index 599
Preface
Physical and natural phenomena depend on a complex array of factors. The sociol-
ogist or psychologist who studies group behavior, the economist who endeavors
to understand the vagaries of a nation's employment cycles, the physicist who
observes the trajectory of a particle or planet, or indeed anyone who seeks to
understand geometry in two, three, or more dimensions recognizes the need to
analyze changing quantities that depend on more than a single variable. Vec-
tor calculus is the essential mathematical tool for such analysis. Moreover, it
is an exciting and beautiful subject in its own right, a true adventure in many
dimensions.
The only technical prerequisite for this text, which is intended for a
sophomore-level course in multivariable calculus, is a standard course in the cal-
culus of functions of one variable. In particular, the necessary matrix arithmetic
and algebra (not linear algebra) are developed as needed. Although the mathe-
matical background assumed is not exceptional, the reader will still be challenged
in places.
My own objectives in writing the book are simple ones: to develop in students
a sound conceptual grasp of vector calculus and to help them begin the transition
from first-year calculus to more advanced technical mathematics. I maintain that
the first goal can be met, at least in part, through the use of vector and matrix
notation, so that many results, especially those of differential calculus, can be
stated with reasonable levels of clarity and generality. Properly described, results
in the calculus of several variables can look quite similar to those of the calculus
of one variable. Reasoning by analogy will thus be an important pedagogical tool.
I also believe that a conceptual understanding of mathematics can be obtained
through the development of a good geometric intuition. Although I state many
results in the case of n variables (where n is arbitrary), I recognize that the most
important and motivational examples usually arise for functions of two and three
variables, so these concrete and visual situations are emphasized to explicate the
general theory. Vector calculus is in many ways an ideal subject for students
to begin exploration of the interrelations among analysis, geometry, and matrix
algebra.
Multivariable calculus, for many students, represents the beginning of signif-
icant mathematical maturation. Consequently, I have written a rather expansive
text so that they can see that there is a story behind the results, techniques, and
examples — that the subject coheres and that this coherence is important for prob-
lem solving. To indicate some of the power of the methods introduced, a number
of topics, not always discussed very fully in a first multivariable calculus course,
are treated here in some detail:
• an early introduction of cylindrical and spherical coordinates (§1.7);
• the use of vector techniques to derive Kepler's laws of planetary motion
(§3.1);
• the elementary differential geometry of curves in R3, including discussion
of curvature, torsion, and the Frenet-Serret formulas for the moving frame
(§3.2);
• Taylor's formula for functions of several variables (§4.1);
X Preface
• the use of the Hessian matrix to determine the nature (as local extrema) of
critical points of functions of n variables (§4.2 and §4.3);
• an extended discussion of the change of variables formula in double and triple
integrals (§5.5);
• applications of vector analysis to physics (§7.4);
• an introduction to differential forms and the generalized Stokes's theorem
(Chapter 8).
Included are a number of proofs of important results. The more techni-
cal proofs are collected as addenda at the ends of the appropriate sections so
as not to disrupt the main conceptual flow and to allow for greater flexibility
of use by the instructor and student. Nonetheless, some proofs (or sketches of
proofs) embody such central ideas that they are included in the main body of the
text.
New in the Fourth Edition
I have retained the overall structure and tone of prior editions. New features in
this edition include the following:
• 210 additional exercises, at all levels;
• a new, optional section (§5.7) on numerical methods for approximating
multiple integrals;
• reorganization of the material on Newton's method for approximating
solutions to systems of n equations in n unknowns to its own (optional)
section (§2.7);
• new proofs in Chapter 2 of limit properties (in §2.2) and of the general
multivariable chain rule (Theorem 5.3 in §2.5);
• proofs of both single-variable and multivariable versions of Taylor's theorem
in §4.1;
• various additional refinements and clarifications throughout the text,
including many new and revised examples and explanations;
• new Microsoft® PowerPoint® files and Wolfram Mathematica® notebooks
that coordinate with the text and that instructors may use in their teaching
(see "Ancillary Materials" below).
How to Use This Book
There is more material in this book than can be covered comfortably during a single
semester. Hence, the instructor will wish to eliminate some topics or subtopics — or
to abbreviate the rather leisurely presentations of limits and differentiability. Since
I frequently find myself without the time to treat surface integrals in detail, I have
separated all material concerning parametrized surfaces, surface integrals, and
Stokes's and Gauss's theorems (Chapter 7), from that concerning line integrals
and Green's theorem (Chapter 6). In particular, in a one-semester course for
students having little or no experience with vectors or matrices, instructors can
probably expect to cover most of the material in Chapters 1-6, although no doubt
it will be necessary to omit some of the optional subsections and to downplay
Preface xi
many of the proofs of results. A rough outline for such a course, allowing for
some instructor discretion, could be the following:
Chapter 1 8-9 lectures
If students have a richer background (so that much of the material in Chapter 1
can be left largely to them to read on their own), then it should be possible to treat
a good portion of Chapter 7 as well. For a two-quarter or two-semester course,
it should be possible to work through the entire book with reasonable care and
rigor, although coverage of Chapter 8 should depend on students' exposure to
introductory linear algebra, as somewhat more sophistication is assumed there.
The exercises vary from relatively routine computations to more challenging
and provocative problems, generally (but not invariably) increasing in difficulty
within each section. In a number of instances, groups of problems serve to intro-
duce supplementary topics or new applications. Each chapter concludes with a
set of miscellaneous exercises that both review and extend the ideas introduced
in the chapter.
A word about the use of technology. The text was written without reference
to any particular computer software or graphing calculator. Most of the exercises
can be solved by hand, although there is no reason not to turn over some of the
more tedious calculations to a computer. Those exercises that require a computer
for computational or graphical purposes are marked with the symbol ^ and
should be amenable to software such as Mathematical ', Maple®, or MATLAB.
Ancillary Materials
In addition to this text a Student Solutions Manual is available. An Instructor's
Solutions Manual, containing complete solutions to all of the exercises, is
available to course instructors from the Pearson Instructor Resource Center
(www.pearsonhighered.com/irc), as are many Microsoft® PowerPoint® files and
Wolfram Mathematica® notebooks that can be adapted for classroom use. The
reader can find errata for the text and accompanying solutions manuals at the
following address:
www.oberlin.edu/math/faculty/colley/VCErrata.html
Acknowledgments
I am very grateful to many individuals for sharing with me their thoughts and ideas
about multivariable calculus. I would like to express particular appreciation to my
Oberlin colleagues (past and present) Bob Geitz, Kevin Hartshorn, Michael Henle
(who, among other things, carefully read the draft of Chapter 8), Gary Kennedy,
Dan King, Greg Quenell, Michael Raney, Daniel Steinberg, Daniel Styer, Richard
Vale, Jim Walsh, and Elizabeth Wilmer for their conversations with me. I am also
grateful to John Alongi, Northwestern University; Matthew Conner, University
of California, Davis; Henry C. King, University of Maryland; Stephen B. Maurer,
Chapter 2
Chapter 3
Chapter 4
Chapter 5
Chapter 6
9 lectures
4- 5 lectures
5- 6 lectures
8 lectures
4 lectures
38^11 lectures
Swarthmore College; Karen Saxe, Macalester College; David Singer, Case West-
ern Reserve University; and Mark R. Treuden, University of Wisconsin at Stevens
Point, for their helpful comments. Several colleagues reviewed various versions
of the manuscript, and I am happy to acknowledge their efforts and many fine
suggestions. In particular, for the first three editions, I thank the following re-
viewers:
Raymond J. Cannon, Baylor University;
Richard D. Carmichael, Wake Forest University;
Stanley Chang, Wellesley College;
Marcel A. F. Deruaz, University of Ottawa (now emeritus);
Krzysztof Galicki, University of New Mexico (deceased);
Dmitry Gokhman, University of Texas at San Antonio;
Isom H. Herron, Rensselaer Polytechnic Institute;
Ashwani K. Kapila, Rensselaer Polytechnic Institute;
Christopher C. Leary, State University of New York, College at Geneseo;
David C. Minda, University of Cincinnati;
Jeffrey Morgan, University of Houston;
Monika Nitsche, University of New Mexico;
Jeffrey L. Nunemacher, Ohio Wesleyan University;
Gabriel Prajitura, State University of New York, College at Brockport;
Florin Pop, Wagner College;
John T. Scheick, The Ohio State University (now emeritus);
Mark Schwartz, Ohio Wesleyan University;
Leonard M. Smiley, University of Alaska, Anchorage;
Theodore B. Stanford, New Mexico State University;
James Stasheff, University of North Carolina at Chapel Hill (now emeritus);
Saleem Watson,California State University, Long Beach;
Floyd L. Williams, University of Massachusetts, Amherst (now emeritus).
For the fourth edition, I thank:
Justin Corvino, Lafayette College;
Carrie Finch, Washington and Lee University;
Soomin Kim, Johns Hopkins University;
Tanya Leise, Amherst College;
Bryan Mosher, University of Minnesota.
Many people at Oberlin College have been of invaluable assistance through-
out the production of all the editions of Vector Calculus. I would especially like
to thank Ben Miller for his hard work establishing the format for the initial drafts
and Stephen Kasperick-Postellon for his manifold contributions to the typeset-
ting, indexing, proofreading, and friendly critiquing of the original manuscript. I
am very grateful to Linda Miller and Michael Bastedo for their numerous typo-
graphical contributions and to Catherine Murillo for her help with any number of
tasks. Thanks also go to Joshua Davis and Joaquin Espinoza Goodman for their
assistance with proofreading. Without the efforts of these individuals, this project
might never have come to fruition.
The various editorial and production staff members have been most kind and
helpful to me. For the first three editions, I would like to express my appreciation
to my editor, George Lobell, and his editorial assistants Gale Epps, Melanie
Van Benthuysen, and Jennifer Urban; to production editors Nicholas Romanelli,
Barbara Mack, and Debbie Ryan at Prentice Hall, and Lori Hazzard at Interactive
Composition Corporation; to Ron Weickart and the staff at Network Graphics
Preface xiii
for their fine rendering of the figures, and to Tom Benfatti of Prentice Hall for
additional efforts with the figures; and to Dennis Kletzing for his careful and
enthusiastic composition work. For this edition, it is a pleasure to acknowledge
my upbeat editor, Caroline Celano, and her assistant, Brandon Rawnsley; they
have made this new edition fun to do. In addition, I am most grateful to Beth
Houston, my production manager at Pearson, Jogender Taneja and the staff at
Aptara, Inc., Donna Mulder, Roger Lipsett, and Thomas Wegleitner.
Finally, I thank the many Oberlin students who had the patience to listen to
me lecture and who inspired me to write and improve this volume.
SJC
sjcolley@math.oberlin.edu
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To the Student:
Some Preliminary
Notation
Here are the ideas that you need to keep in mind as you read this book and learn
vector calculus.
Given two sets A and B, I assume that you are familiar with the notation
A U B for the union of A and B — those elements that are in either A or B (or
both):
AU B = {x\x € Aorx € B}.
Similarly, A n B is used to denote the intersection of A and B — those elements
that are in both A and B :
A(~)B = {x\x<=A andx e B}.
The notation A c B, or A c B, indicates that A is a subset of B (possibly empty
or equal to B).
One-dimensional space (also called the real line or R) is just a straight line.
We put real number coordinates on this line by placing negative numbers on the
left and positive numbers on the right. (See Figure 1.)
Two-dimensional space, denoted R2, is the familiar Cartesian plane. If we
construct two perpendicular lines (the x- and y-coordinate axes), set the origin
as the point of intersection of the axes, and establish numerical scales on these
lines, then we may locate a point in R2 by giving an ordered pair of numbers (x , y),
the coordinates of the point. Note that the coordinate axes divide the plane into
four quadrants. (See Figure 2.)
Three-dimensional space, denoted R3, requires three mutually perpendicular
coordinate axes (called the x-, y- and z-axes) that meet in a single point (called
the origin) in order to locate an arbitrary point. Analogous to the case of R2, if we
establish scales on the axes, then we can locate a point in R3 by giving an ordered
triple of numbers (x, y, z). The coordinate axes divide three-dimensional space
into eight octants. It takes some practice to get your sense of perspective correct
when sketching points in R3. (See Figure 3.) Sometimes we draw the coordinate
axes in R3 in different orientations in order to get a better view of things. However,
we always maintain the axes in a right-handed configuration. This means that
if you curl the fingers of your right hand from the positive x-axis to the positive
y-axis, then your thumb will point along the positive z-axis. (See Figure 4.)
Although you need to recall particular techniques and methods from the
calculus you have already learned here are some of the more important concepts
to keep in mind: Given a function f(x), the derivative f'(x) is the limit (if it exists)
of the difference quotient of the function:
f (x ) = hm .
J v ft^o h
XVI To the Student: Some Preliminary Notation
(2,4,5)
Figure 3 Three-dimensional
space R3 . Selected points are
graphed.
Figure 4 The x-, y-, and z-axes in R3 are always
drawn in a right-handed configuration.
The significance of the derivative /'(xo) is that it measures the slope of the line
tangent to the graph of / at the point (xo, /(xo)). (See Figure 5.) The derivative
may also be considered to give the instantaneous rate of change of / at x = Xq.
We also denote the derivative f'(x) by df/dx.
The definite integral fb f(x) dx of / on the closed interval [a , b] is the limit
(provided it exists) of the so-called Riemann sums of /:
Figure 5 The derivative f'(xo) is
the slope of the tangent line to
y = f (x) at (x0, /(x0)).
f
f(x)dx = lim
all Ax,-*0
£/(x*)Ax,.
(=1
Here a = xo < x\ < xi < • • • < x„ = b denotes a partition of [a, b] into subin-
tervals [x(_i, x, ], the symbol A (the length of the subinterval), and
x* denotes any point in [x/_i , x,-]. If f(x) > 0 on [a, b], then each term f(x*)Axj
in the Riemann sum is the area of a rectangle related to the graph of /. The
Riemann sum YH=i f(x*)^-x< mus approximates the total area under the graph
of / between x = a and x = b. (See Figure 6.)
Figure 6 If f(x) > 0 on [a, b], then the Riemann sum
approximates the area under y = f(x) by giving the sum
of areas of rectangles.
To the Student: Some Preliminary Notation XVII
The definite integral fb f(x)dx, if it exists, is taken to represent the area
under y = f(x) between x = a and x = b. (See Figure 7.)
The derivative and the definite integral are connected by an elegant result
known as the fundamental theorem of calculus. Let f(x) be a continuous func-
tion of one variable, and let F(x) be such that F'(x) = f(x). (The function F is
called an antiderivative of /.) Then
I
b
f(x)dx = F(b)-F(a);
d fx
2. — / f(t)dt = f{x).
dx J a
Finally, the end of an example is denoted by the symbol ♦ and the end of a
proof by the symbol ■.
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Vector Calculus
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Vectors
1.1 Vectors in Two and Three
Dimensions
1 .2 More About Vectors
1 .3 The Dot Product
1 .4 The Cross Product
1.5 Equations for Planes;
Distance Problems
1 .6 Some n-dimensional
Geometry
1.7 New Coordinate Systems
True/False Exercises for
Chapter 1
Miscellaneous Exercises for
Chapter 1
1 .1 Vectors in Two and Three Dimensions
For your study of the calculus of several variables, the notion of a vector is
fundamental. As is the case for many of the concepts we shall explore, there are
both algebraic and geometric points of view. You should become comfortable
with both perspectives in order to solve problems effectively and to build on your
basic understanding of the subject.
Vectors in R2 and R3: The Algebraic Notion
DEFINITION 1.1 A vector in R2 is simply an ordered pair of real numbers.
That is, a vector in R2 may be written as
(aua2) (e.g., (1,2) or (n, 17)).
Similarly, a vector in R3 is simply an ordered triple of real numbers. That is,
a vector in R3 may be written as
(a\,a2, a3) (e.g., (jt, e, <s/2)).
To emphasize that we want to consider the pair or triple of numbers as a
single unit, we will use boldface letters; hence a = (a\, a2) or a = (ai, a%, a{)
will be our standard notation for vectors in R2 or R3 . Whether we mean that a is a
vector in R2 or in R3 will be clear from context (or else won't be important to the
discussion). When doing handwritten work, it is difficult to "boldface" anything,
so you'll want to put an arrow over the letter. Thus, a will mean the same thing
as a. Whatever notation you decide to use, it's important that you distinguish the
vector a (or a) from the single real number a. To contrast them with vectors, we
will also refer to single real numbers as scalars.
In order to do anything interesting with vectors, it's necessary to develop
some arithmetic operations for working with them. Before doing this, however,
we need to know when two vectors are equal.
DEFINITION 1.2 Two vectors a = (ai,a2) and b = (bi,b2) in R2 are
equal if their corresponding components are equal, that is, if a\ = b\ and
a2 = b2. The same definition holds for vectors in R3: a = (a\, a2, a^) and
b = (b\, b2,b-i) are equal if their corresponding components are equal, that
is, if a\ =b\,a2 = b2, and a^ = b^.
2 Chapter 1 ] Vectors
EXAMPLE 1 The vectors a = (1, 2) and b = (§, f ) are equal in R2, but c =
(1, 2, 3) and d = (2, 3, 1) are not equal in R3. ♦
Next, we discuss the operations of vector addition and scalar multiplication.
We'll do this by considering vectors in R3 only; exactly the same remarks will
hold for vectors in R2 if we simply ignore the last component.
DEFINITION 1.3 (Vector Addition) Let a = (a\, a2, 03) and b = (b\,
b2,b^) be two vectors in R3. Then the vector sum a + b is the vector in R3
obtained via componentwise addition: a 4- b = (a\ + b\, a2 + b2, 03 + ^3).
EXAMPLE 2 We have (0, 1,3) + (7, -2, 10) = (7, - 1, 13) and (in R2):
(1, 1) + (7T, V2) = (1 + 7T, 1 + V2). ♦
Properties of vector addition. We have
1. a + b = b + a for all a, b in R3 (commutativity);
2. a + (b + c) = (a + b) + c for all a, b, c in R3 (associativity);
3. a special vector, denoted 0 (and called the zero vector), with the property
that a + 0 = a for all a in R3 .
These three properties require proofs, which, like most facts involving the al-
gebra of vectors, can be obtained by explicitly writing out the vector components.
For example, for property 1, we have that if
a = {a\,a2,a{) and b = {b\, b2, ^3),
then
a + b = (czi + b\ , a2 + b2, «3 + h)
= (b\ + c/i , b2 + a2, b-i + a3)
= b + a,
since real number addition is commutative. For property 3, the "special vector"
is just the vector whose components are all zero: 0 = (0, 0, 0). It's then easy to
check that property 3 holds by writing out components. Similarly for property 2,
so we leave the details as exercises.
DEFINITION 1 .4 (Scalar Multiplication) Let a = (a\ , a2, 03) be a vec-
tor in R3 and let k e R be a scalar (real number). Then the scalar prod-
uct ka is the vector in R3 given by multiplying each component of a by
k: ka = (ka\ , ka2, ka{).
EXAMPLE 3 Ifa = (2,0, V2)and^ = 7,thenyta = (14, 0, 7V2). ♦
The results that follow are not difficult to check — just write out the vector
components.
1.1 | Vectors in Two and Three Dimensions
Properties of scalar multiplication. For all vectors a and b in R3 (or R2)
and scalars k and / in R, we have
1. (k + /)a = ka + la (distributivity);
2. k(a + b) = ka + kb (distributivity);
3. k(la) = (kl)a = l(ka).
It is worth remarking that none of these definitions or properties really de-
pends on dimension, that is, on the number of components. Therefore we could
have introduced the algebraic concept of a vector in R" as an ordered n-tuple
(fli, d2, . . . , an) of real numbers and defined addition and scalar multiplication
in a way analogous to what we did for R2 and R3. Think about what such a
generalization means. We will discuss some of the technicalities involved in § 1 .6.
Vectors in R2 and R3: The Geometric Notion
Although the algebra of vectors is certainly important and you should become
adept at working algebraically, the formal definitions and properties tend to present
a rather sterile picture of vectors. A better motivation for the definitions just given
comes from geometry. We explore this geometry now. First of all, the fact that
a vector a in R2 is a pair of real numbers (a\ , ai) should make you think of the
coordinates of a point in R2. (See Figure 1.1.) Similarly, if a e R3, then a may
be written as (a\, 02, as), and this triple of numbers may be thought of as the
coordinates of a point in R3. (See Figure 1.2.)
All of this is fine, but the results of performing vector addition or scalar mul-
tiplication don't have very interesting or meaningful geometric interpretations in
terms of points. As we shall see, it is better to visualize a vector in R2 or R3 as an
arrow that begins at the origin and ends at the point. (See Figure 1.3.) Such a depic-
tion is often referred to as the position vector of the point (a\ , ai) or {a\ , #2, ai).
If you've studied vectors in physics, you have heard them described as objects
having "magnitude and direction." Figure 1 .3 demonstrates this concept, provided
that we take "magnitude" to mean "length of the arrow" and "direction" to be the
orientation or sense of the arrow. (Note: There is an exception to this approach,
namely, the zero vector. The zero vector just sits at the origin, like a point, and has
no magnitude and therefore, an indeterminate direction. This exception will not
pose much difficulty.) However, in physics, one doesn't demand that all vectors
4 Chapter 1 | Vectors
be represented by arrows having their tails bound to the origin. One is free to
"parallel translate" vectors throughout R2 and R3. That is, one may represent
the vector a = (a\ , a2, a^) by an arrow with its tail at the origin (and its head at
(fli, a2, 03)) or with its tail at any other point, so long as the length and sense of
the arrow are not disturbed. (See Figure 1 .4.) For example, if we wish to represent
a by an arrow with its tail at the point (jci, x2, ^3), then the head of the arrow
would be at the point {x\ + ci\ , x2 + a2, X3 + 03). (See Figure 1.5.)
Figure 1 .4 Each arrow is a
parallel translate of the position
vector of the point (ai , 02, 03) and
represents the same vector.
(^i + flj, x2 + a2, x3 + a3)
(xhx2,x3)
Figure 1 .5 The vector
a = (a\ , ci2, 03) represented by an
arrow with tail at the point
(xi,x2, x3).
Figure 1 .6 The vector
a + b may be represented
by an arrow whose tail is at
the tail of a and whose head
is at the head of b.
With this geometric description of vectors, vector addition can be visualized
in two ways. The first is often referred to as the "head-to-tail" method for adding
vectors. Draw the two vectors a and b to be added so that the tail of one of the
vectors, say b, is at the head of the other. Then the vector sum a + b may be
represented by an arrow whose tail is at the tail of a and whose head is at the head
of b. (See Figure 1.6.) Note that it is not immediately obvious that a + b = b + a
from this construction!
The second way to visualize vector addition is according to the so-called
parallelogram law: If a and b are nonparallel vectors drawn with their tails ema-
nating from the same point, then a + b may be represented by the arrow (with its
tail at the common initial point of a and b) that runs along a diagonal of the paral-
lelogram determined by a and b (Figure 1.7). The parallelogram law is completely
consistent with the head-to-tail method. To see why, just parallel translate b to the
opposite side of the parallelogram. Then the diagonal just described is the result of
adding a and (the translate of) b, using the head-to-tail method. (See Figure 1.8.)
We still should check that these geometric constructions agree with our alge-
braic definition. For simplicity, we'll work in R2. Let a = {a\ , a2) and b = (b\ , b2)
as usual. Then the arrow obtained from the parallelogram law addition of a and
b is the one whose tail is at the origin O and whose head is at the point P in
Figure 1.9. If we parallel translate b so that its tail is at the head of a, then it is
immediate that the coordinates of P must be (a\ + b\,a2 + b2), as desired.
Scalar multiplication is easier to visualize: The vector ka. may be represented
by an arrow whose length is \k\ times the length of a and whose direction is the
same as that of a when k > 0 and the opposite when k < 0. (See Figure 1.10.)
It is now a simple matter to obtain a geometric depiction of the difference
between two vectors. (See Figure 1.11.) The difference a — b is nothing more
1.1 | Vectors in Two and Three Dimensions
a + b /
/
/
i
a + b
(translated)
Figure 1 .7 The vector
a + b may be represented
by the arrow that runs along
the diagonal of the
parallelogram determined
by a and b.
Figure 1 .8 The equivalence of the
parallelogram law and the
head-to-tail methods of vector
addition.
Figure 1.11 The
geometry of vector
subtraction. The vector c
is such that b + c = a.
Hence, c = a — b.
2<
s_ —
b / / b
V v
Figure 1 .9 The point P has coordinates
(«i + b\, a2 + b2).
Figure 1.10 Visualization of
scalar multiplication.
than a + (— b) (where — b means the scalar —1 times the vector b). The vector
a — b may be represented by an arrow pointing from the head of b toward the
head of a; such an arrow is also a diagonal of the parallelogram determined by a
and b. ( As we have seen, the other diagonal can be used to represent a + b.)
Here is a construction that will be useful to us from time to time.
DEFINITION 1 .5 Given two points Pi(xi , yi , zi) and P2(x2, y2, z2) in R3,
the displacement vector from P\ to P2 is
P\A = (xi - x\, yi - yi, zi - zi).
This construction is not hard to understand if we consider Figure 1.12. Given
the points P\ and P2, draw the corresponding position vectors OP\ and OP2-
Then we see that P\P2 is precisely OP2—OP\. An analogous definition may
~y be made for R2.
/ In your study of the calculus of one variable, you no doubt used the notions of
derivatives and integrals to look at such physical concepts as velocity, acceleration,
Figure 1 .1 2 The displacement force, etc. The main drawback of the work you did was that the techniques involved
vector P\P2, represented by the allowed you to study only rectilinear, or straight-line, activity. Intuitively, we all
arrow from Pi to P2, is the understand that motion in the plane or in space is more complicated than straight-
difference between the position line motion. Because vectors possess direction as well as magnitude, they are
vectors of these two points. ideally suited for two- and three-dimensional dynamical problems.
Chapter 1 | Vectors
Figure 1.13 After t seconds, the
point starting at a, with velocity v,
moves to a + t\.
Vj ship
(with respect
to still water)
, current
Net velocity
Figure 1.14 The length of Vi is
15, and the length of V2 is 5V2.
For example, suppose a particle in space is at the point (ai, a.2, a{) (with
respect to some appropriate coordinate system). Then it has position vector a =
(a\, a2, 03). If the particle travels with constant velocity v = (uj, i>2, V3) for /
seconds, then the particle's displacement from its original position is rv, and its
new coordinate position is a + fv. (See Figure 1.13.)
EXAMPLE 4 If a spaceship is at position (100, 3, 700) and is traveling with
velocity (7, —10,25) (meaning that the ship travels 7 mi/sec in the positive
x-direction, 10 mi/sec in the negative y-direction, and 25 mi/sec in the positive
z-direction), then after 20 seconds, the ship will be at position
(100, 3, 700) + 20(7, -10, 25) = (240, -197, 1200),
and the displacement from the initial position is (140, —200, 500). ♦
EXAMPLE 5 The S.S. Calculus is cruising due south at a rate of 15 knots
(nautical miles per hour) with respect to still water. However, there is also a
current of 5^2 knots southeast. What is the total velocity of the ship? If the ship
is initially at the origin and a lobster pot is at position (20, —79), will the ship
collide with the lobster pot?
Since velocities are vectors, the total velocity of the ship is Vi + V2, where vi is
the velocity of the ship with respect to still water and V2 is the southeast-pointing
velocity of the current. Figure 1.14 makes it fairly straightforward to compute
these velocities. We have that vi = (0, —15). Since \2 points southeastward, its
direction must be along the line v = — x. Therefore, V2 can be written as V2 =
(v, —v), where v is a positive real number. By the Pythagorean theorem, if the
length of \2 is 5\/2, then we must have v2 + (— v)2 = (5>/2)2 or 2v2 = 50, so
that v = 5. Thus, v2 = (5, —5), and hence, the net velocity is
(0, -15) + (5, -5) = (5, -20).
After 4 hours, therefore, the ship will be at position
(0,0) + 4(5, -20) = (20, -80)
and thus will miss the lobster pot. ♦
EXAMPLE 6 The theory behind the venerable martial art of judo is an excel-
lent example of vector addition. If two people, one relatively strong and the other
relatively weak, have a shoving match, it is clear who will prevail. For example,
someone pushing one way with 200 lb of force will certainly succeed in overpow-
ering another pushing the opposite way with 1 00 lb of force. Indeed, as Figure 1.15
shows, the net force will be 100 lb in the direction in which the stronger person
is pushing.
1001b
2001b
1001b
200 lb
2001b
Figure 1.16 Vector addition in
judo.
Figure 1 .1 5 A relatively strong person pushing with a
force of 200 lb can quickly subdue a relatively weak one
pushing with only 100 lb of force.
Dr. Jigoro Kano, the founder of judo, realized (though he never expressed
his idea in these terms) that this sort of vector addition favors the strong over the
weak. However, if the weaker participant applies his or her 100 lb of force in a
direction only slightly different from that of the stronger, he or she will effect a
vector sum of length large enough to surprise the opponent. (See Figure 1.16.)
1.1 | Exercises
This is the basis for essentially all of the throws of judo and why judo is described
as the art of "using a person's strength against himself or herself." In fact, the
word "judo" means "the giving way." One "gives in" to the strength of another by
attempting only to redirect his or her force rather than to oppose it. ♦
1.1 Exercises
1 . Sketch the following vectors in R2 :
(a) (2,1) (b)(3,3) (c)(-l,2)
2. Sketch the following vectors in R3 :
(a) (1,2,3) (b) (-2,0,2) (c) (2, -3, 1)
3. Perform the indicated algebraic operations. Express
your answers in the form of a single vector a = (a \ , 02)
inR2.
(a) (3,1) + (-1,7)
(b) -2(8, 12)
(c) (8,9) + 3(-l,2)
(d) (1, 1) + 5(2, 6) -3(10, 2)
(e) (8, 10) + 3 ((8, -2) - 2(4, 5))
4. Perform the indicated algebraic operations. Express
your answers in the form of a single vector a =
(ai , «2> ai) in R3-
(a) (2, 1,2) + (-3, 9, 7)
(b) 1(8,4, 1) + 2(5,-7, ±)
(c) -2 ((2,0, l)-6(i,-4, 1))
5. Graph the vectors a = (1, 2), b = (-2, 5), and a +
b = ( 1 , 2) + (—2, 5), using both the parallelogram law
and the head-to-tail method.
6. Graph the vectors a = (3, 2) and b = (— 1, 1). Also
calculate and graph a — b, ia, and a + 2b.
7. Let A be the point with coordinates (1,0, 2), let B be
the point with coordinates (—3, 3, 1), and let C be the
point with coordinates (2, 1,5).
(a) Describe the vectors AB and BA.
(b) Describe the vectors AC, B~C, and AC + CB.
(c) Explain, with pictures, why AC + CB = AB.
8. Graph (1,2, 1 ) and (0,-2, 3), and calculate and graph
(1, 2, 1) + (0, -2, 3), -1(1,2, 1), and 4(1, 2, 1).
9. If (-12, 9, z) + (x, 7, -3) = (2, y, 5), what are x, y,
andz?
10. What is the length (magnitude) of the vector (3, 1)?
(Hint: A diagram will help.)
11. Sketch the vectors a = (l,2)andb = (5, 10). Explain
why a and b point in the same direction.
12. Sketch the vectors a = (2, -7, 8) and b=(-l,
j,— 4). Explain why a and b point in opposite
directions.
13. How would you add the vectors (1,2,3,4) and
(5,-1, 2, 0) in R4? What should 2(7, 6, -3, 1) be? In
general, suppose that
a = (a\ , <22, • ■ ■ , a„) and b = (bi, b%, ■ ■ ■ , bn)
are two vectors in R" and k e R is a scalar. Then how
would you define a + b and fca?
14. Find the displacement vectors from Pi to P2, where Pi
and P2 are the points given. Sketch Pi, P2, and P1P2.
(a) Pi(l,0,2),P2(2, 1,7)
(b) Pi(l,6,-l),P2(0,4,2)
(c) Pi(0,4,2),P2(l,6,-l)
(d) Pi(3, 1),P2(2,-1)
15. Let Pi(2, 5, -1, 6) and P2(3, 1, -2, 7) be two points
in R4. How would you define and calculate the dis-
placement vector from Pi to P2? (See Exercise 13.)
16. If A is the point in R3 with coordinates (2, 5, —6) and
the displacement vector from A to a second point B is
(12, —3, 7), what are the coordinates of Bl
1 7. Suppose that you and your friend are in New York talk-
ing on cellular phones. You inform each other of your
own displacement vectors from the Empire State Build-
ing to your current position. Explain how you can use
this information to determine the displacement vector
from you to your friend.
1 8. Give the details of the proofs of properties 2 and 3 of
vector addition given in this section.
1 9. Prove the properties of scalar multiplication given in
this section.
20. (a) If a is a vector in R2 or R3, what is 0a? Prove your
answer.
(b) If a is a vector in R2 or R3, what is la? Prove your
answer.
21. (a) Let a = (2, 0) and b = (1, 1). For 0 < s < 1 and
0 < t < 1, consider the vector x = sa + tb. Ex-
plain why the vector x lies in the parallelogram
8 Chapter 1 ] Vectors
determined by a and b. (Hint: It may help to draw
a picture.)
(b) Now suppose that a = (2, 2, 1) and b = (0, 3, 2).
Describe the set of vectors {x = sa + ?b | 0 < .y <
1, 0 < t < 1}.
22. Let a = {a\, ai, 03) andb = (b\, bi, bi) be two nonzero
vectors such that b / ka. Use vectors to describe
the set of points inside the parallelogram with vertex
Po(xo, yo, zo) and whose adjacent sides are parallel to
a and b and have the same lengths as a and b. (See
Figure 1.17.) (Hint: If P(x, y, z) is a point in the par-
allelogram, describe OP, the position vector of P.)
z
X
Figure 1.17 Figure for Exercise 22.
23. A flea falls onto marked graph paper at the point (3,2).
She begins moving from that point with velocity vector
v = (—1, —2) (i.e., she moves 1 graph paper unit per
minute in the negative x -direction and 2 graph paper
units per minute in the negative y-direction).
(a) What is the speed of the flea?
(b) Where is the flea after 3 minutes?
(c) How long does it take the flea to get to the point
(-4, -12)?
(d) Does the flea reach the point (—13, —27)? Why or
why not?
24. A plane takes off from an airport with velocity vector
(50, 100, 4). Assume that the units are miles per hour,
that the positive a- -axis points east, and that the positive
y-axis points north.
(a) How fast is the plane climbing vertically at take-
off?
(b) Suppose the airport is located at the origin and a
skyscraper is located 5 miles east and 10 miles
north of the airport. The skyscraper is 1 ,250 feet tall.
When will the plane be directly over the building?
(c) When the plane is over the building, how much
vertical clearance is there?
25. As mentioned in the text, physical forces (e.g., gravity)
are quantities possessing both magnitude and direction
and therefore can be represented by vectors . If an obj ect
has more than one force acting on it, then the resul-
tant (or net) force can be represented by the sum of
the individual force vectors. Suppose that two forces,
F] = (2, 7, -1) and F2 = (3, -2, 5), act on an object.
(a) What is the resultant force of Fj and F2?
(b) What force F3 is needed to counteract these forces
(i.e., so that no net force results and the object
remains at rest)?
26. A 50 lb sandbag is suspended by two ropes. Suppose
that a three-dimensional coordinate system is intro-
duced so that the sandbag is at the origin and the ropes
are anchored at the points (0, —2, 1) and (0, 2, 1).
(a) Assuming that the force due to gravity points par-
allel to the vector (0, 0,-1), give a vector F that
describes this gravitational force.
(b) Now, use vectors to describe the forces along each
of the two ropes. Use symmetry considerations and
draw a figure of the situation.
27. A 10 lb weight is suspended in equilibrium by two
ropes. Assume that the weight is at the point (1, 2, 3)
in a three-dimensional coordinate system, where the
positive z-axis points straight up, perpendicular to the
ground, and that the ropes are anchored at the points
(3,0,4) and (0,3,5). Give vectors Fj and F2 that
describe the forces along the ropes.
1 .2 More About Vectors
The Standard Basis Vectors
In R2, the vectors i = (1, 0) and j = (0, 1) play a special notational role. Any
vector a= (a\, Oq) may be written in terms of i and j via vector addition and
scalar multiplication:
a2) — (fli, 0) + (0, a2) = fli(l, 0) + a2(0, 1) = ax i + a2 j.
(It may be easier to follow this argument by reading it in reverse.) Insofar as nota-
tion goes, the preceding work simply establishes that one can write either (a\ , a2)
1.2 | More About Vectors
j
i
«2j
----^^ a = a{\ + a2
Figure 1.18 Any vector in R2 can be written in terms of i and j.
Figure 1.19 Any vector in R can be written in terms of i, j, and k.
v = 3
Figure 1 .20 In R2, the equation
y = 3 describes a line.
or aji + ci2] to denote the vector a. It's your choice which notation to use (as long
as you're consistent), but the ij-notation is generally useful for emphasizing the
"vector" nature of a, while the coordinate notation is more useful for emphasizing
the "point" nature of a (in the sense of a's role as a possible position vector of
a point). Geometrically, the significance of the standard basis vectors i and j is
that an arbitrary vector a e R2 can be decomposed pictorially into appropriate
vector components along the x- and y-axes, as shown in Figure 1.18.
Exactly the same situation occurs in R3, except that we need three vec-
tors, i = (1 , 0, 0), j = (0, 1 , 0), and k = (0, 0, 1), to form the standard basis. (See
Figure 1.19.) The same argument as the one just given can be used to show that
any vector a = (a\,a2, a^) may also be written as a\ i + aj. j + c?3 k. We shall
use both coordinate and standard basis notation throughout this text.
Figure 1 .21 In R3, the equation
y = 3 describes a plane.
EXAMPLE 1 We may write the vector (1,
(7, n, —3) as 7i + 7ij — 3k.
-2) as i — 2j and the vector
♦
Parametric Equations of Lines
In R2, we know that equations of the form y = mx + b or Ax + By = C describe
straight lines. (See Figure 1.20.) Consequently, one might expect the same sort of
equation to define a line in R3 as well. Consideration of a simple example or two
(such as in Figure 1.21) should convince you that a single such linear equation
describes a plane, not a line. A pair of simultaneous equations in x, y, and z is
required to define a line.
We postpone discussing the derivation of equations for planes until §1.5 and
concentrate here on using vectors to give sets of parametric equations for lines in
R2 orR3 (or even R").
1 0 Chapter 1 | Vectors
t = 3n:/2
Figure 1 .22 The graph of the
parametric equations x = 2 cos t,
y = 2 sin t , 0 < t < 2n .
First, we remark that a curve in the plane may be described analytically
by points (x, y), where x and y are given as functions of a third variable (the
parameter) t. These functions give rise to parametric equations for the curve:
x = fit)
y = g(t) '
EXAMPLE 2 The set of equations
x = 2cosf
0 < t < In
y = 2 sin t
describes a circle of radius 2, since we may check that
x2 + y2 = (2 cos t)2 + (2 sin t)2 = 4.
(See Figure 1.22.)
Figure 1 .23 The line / is the
unique line passing through Pq and
parallel to the vector a.
Parametric equations may be used as readily to describe curves in R3 ; a curve
in R3 is the set of points (x, y, z) whose coordinates x,y, and z are each given by
a function of t:
x = f(t)
y = g(t) ■
z. = hit)
The advantages of using parametric equations are twofold. First, they offer a
uniform way of describing curves in any number of dimensions. (How would
you define parametric equations for a curve in R4? In R128?) Second, they allow
you to get a dynamic sense of a curve if you consider the parameter variable t to
represent time and imagine that a particle is traveling along the curve with time
according to the given parametric equations. You can represent this geometrically
by assigning a "direction" to the curve to signify increasing t. Notice the arrow
in Figure 1.22.
Now, we see how to provide equations for lines. First, convince yourself that
a line in R2 or R3 is uniquely determined by two pieces of geometric information:
(1) a vector whose direction is parallel to that of the line and (2) any particular
point lying on the line — see Figure 1.23. In Figure 1.24, we seek the vector
r = ~OP
between the origin O and an arbitrary point P on the line / (i.e., the position
vector of P(x, y, z))- OP is the vector sum of the position vector b of the given
point P0 (i.e., OPq) and a vector parallel to a. Any vector parallel to a must be a
scalar multiple of a. Letting this scalar be the parameter variable t, we have
r=0~P~ = OP^o + ta,
and we have established the following proposition:
PROPOSITION 2.1 The vector parametric equation for the line through the point
Poib\ , b2, bi), whose position vector is OPq = b = b\i + b2] + Z?3k, and parallel
to a = aii + a2] + aj,k is
Figure 1 .24 The graph of a line
inR3.
r(f) = b + /a.
(1)
1.2 | More About Vectors 11
Expanding formula (1),
r(f) = OP = b\i + b2j + &3k + t{a\\ + a2\ + o^k)
= (ait + h)i + (a2t + b2)j + (a3t + 63)k.
Next, write op as xi + yj + zk so that P has coordinates (x, y, z). Then, ex-
tracting components, we see that the coordinates of P are (a\t + b\, a2t + fe,
+ b$) and our parametric equations are
x = a\t + b\
■ y = a2t + b2 , (2)
z = a3t + Z?3
where f is any real number.
These parametric equations work just as well in R2 (if we ignore the z-
component) or in R" where n is arbitrary. In R", formula (1) remains valid where
we take a = (a,\, a2, . . . , an) and b = (b\, b2, . . . , b„). The resulting parametric
equations are
x\ = ci\t + b\
x2 = a2t + b2
xn — cin t -\- bn
Figure 1.25 Finding equations
for a line through two points in
Example 4.
EXAMPLE 3 To find the parametric equations of the line through (1, —2, 3)and
parallel to the vector n\ — 3j + k, we have a = ni — 3j + k and b = i — 2j + 3k
so that formula (1) yields
r(/) = i - 2j + 3k + t(ni - 3j + k)
= ( 1 + jtt)\ + (-2 - 3r)j + (3 + r)k
The parametric equations may be read as
X = 7tt + 1
y = -3t-2 .
vz = t + 3 ♦
EXAMPLE 4 From Euclidean geometry, two distinct points determine a unique
line in R2 or R3 . Let's find the parametric equations of the line through the points
Pq(\, —2, 3) and P\(0, 5, —1). The situation is suggested by Figure 1.25. To use
formula (1), we need to find a vector a parallel to the desired line. The vector with
tail at Pq and head at P\ is such a vector. That is, we may use for a the vector
PoP^ = (0-1,5- (-2), -1 - 3) = -i + 7j - 4k.
For b, the position vector of a particular point on the line, we have the choice
of taking either b = i — 2j + 3k or b = 5j — k. Hence, the equations in (2) yield
parametric equations
x = 1 - t
y = -2 + It
z = 3 - At
or
x = — t
y = 5 + It
z = -1 -At
Chapter 1 | Vectors
In general, given two arbitrary points
Po(a\, a2, a3) and P\(b\, b2, fc3),
the line joining them has vector parametric equation
r(t) = OP0 + tP0Pi.
Equation (3) gives parametric equations
x = a\ + (b\ — a\)t
y = a2 + (b2 - a2)t .
Z = a3 + (b2 - ai)t
Alternatively, in place of equation (3), we could use the vector equation
(3)
(4)
r(r) = OP\ + tP0P\
or perhaps
r(r) = OPx + tP\P0
(5)
(6)
each of which gives rise to somewhat different sets of parametric equations. Again,
we refer you to Figure 1 .25 for an understanding of the vector geometry involved.
Example 4 brings up an important point, namely, that parametric equations
for a line (or, more generally, for any curve) are never unique. In fact, the two
sets of equations calculated in Example 4 are by no means the only ones; we
>
could have taken a = Pi Pq = i — 7j + 4k or any nonzero scalar multiple of
P^p\ for a.
If parametric equations are not determined uniquely, then how can you check
your work? In general, this is not so easy to do, but in the case of lines, there are
two approaches to take. One is to produce two points that lie on the line specified
by the first set of parametric equations and see that these points lie on the line
given by the second set of parametric equations. The other approach is to use the
parametric equations to find what is called the symmetric form of a line in R3 .
From the equations in (2), assuming that each a, is nonzero, one can eliminate
the parameter variable t in each equation to obtain:
x — b\
t =
t =
y - b2
a2
z - h
a 3
The symmetric form is
1.2 | More About Vectors 13
In Example 4, the two sets of parametric equations give rise to corresponding
symmetric forms
x — 1 y + 2 z — 3 , x y — 5 z + 1
■1
and
■1
It's not difficult to see that adding 1 to each "side" of the second symmetric form
yields the first one. In general, symmetric forms for lines can differ only by a
constant term or constant scalar multiples (or both).
The symmetric form is really a set of two simultaneous equations in R3. For
example, the information in (7) can also be written as
b\ y - b2
x
fll
x — b\
y
fl3
This illustrates that we require two "scalar" equations in x, y, and z to describe a
line in R3, although a single vector parametric equation, formula (1), is sufficient.
The next two examples illustrate how to use parametric equations for lines to
identify the intersection of a line and a plane or of two lines.
EXAMPLE 5 We find where the line with parametric equations
x = t + 5
y
-2t - 4
3t + 7
intersects the plane 3x + 2y — lz = 2.
To locate the point of intersection, we must find what value of the parameter t
gives a point on the line that also lies in the plane. This is readily accomplished by
substituting the parametric values for x, y, and z from the line into the equation
for the plane
3(r + 5) + 2(-2t - 4) - 7(3f + 7) = 2. (8)
Solving equation (8) for t, we find that t = —2. Setting t equal to —2 in the
parametric equations for the line yields the point (3,0, 1), which, indeed lies in
the plane as well. ♦
EXAMPLE 6 We determine whether and where the two lines
3t - 3
and \ y = t
x — t + 1
y = 5t + 6
z = -2t
x
y
z = t+ 1
intersect.
The lines intersect provided that there is a specific value t\ for the parameter
of the first line and a value h for the parameter of the second line that generate the
same point. In other words, we must be able to find t\ and t2 so that, by equating
the respective parametric expressions for x, y, and z„ we have
h + 1 = 3f2 -
5fi + 6 = t2
-2t{ = t2 + 1
3
(9)
1 4 Chapter 1 | Vectors
The last two equations of (9) yield
t2 = 5?i + 6 = -2*i - 1 => fi = — 1.
Using t\ = — 1 in the second equation of (9), we find that t2 = 1 . Note that the
values t\ = — 1 and ?2 = 1 also satisfy the first equation of (9); therefore, we have
solved the system. Setting t = — 1 in the set of parametric equations for the first
line gives the desired intersection point, namely, (0, 1,2). ♦
Parametric Equations in General
Vector geometry makes it relatively easy to find parametric equations for a variety
of curves. We provide two examples.
EXAMPLE 7 If a wheel rolls along a flat surface without slipping, a point on
the rim of the wheel traces a curve called a cycloid, as shown in Figure 1.26.
Figure 1 .27 The result of the
wheel in Figure 1 .26 rolling
through a central angle of t .
D
3x12 - A
A
Figure 1 .28 AP with its tail at
the origin.
Figure 1 .26 The graph of a cycloid.
Suppose that the wheel has radius a and that coordinates in R2 are chosen so that
the point of interest on the wheel is initially at the origin. After the wheel has
rolled through a central angle of t radians, the situation is as shown in Figure 1.27.
We seek the vector OP, the position vector of P, in terms of the parameter t.
Evidently, OP = OA + AP, where the point A is the center of the wheel. The
vector O A is not difficult to determine. Its j-component must be a , since the center
of the wheel does not vary vertically. Its i-component must equal the distance the
wheel has rolled; if t is measured in radians, then this distance is at, the length
of the arc of the circle having central angle t. Hence, OA = ati + aj.
The value of vector methods becomes apparent when we determine A~P.
Parallel translate the picture so that A~P has its tail at the origin, as in Figure 1 .28.
From the parametric equations of a circle of radius a,
aP =
a cos
3tt
T
+ a sin
3tt
T
? I j = —a sin t\ — a cos t j,
from the addition formulas for sine and cosine. We conclude that
OP = 0~A + A~P = (ati + aj) + (—a sin ti — a cos t\)
= a(t — sin t )i + a(l — cos t)j,
1.2 | More About Vectors 15
so the parametric equations are
x = a(t — sinf)
y = a(l — cos t) ♦
EXAMPLE 8 If you unwind adhesive tape from a nonrotating circular tape
dispenser so that the unwound tape is held taut and tangent to the dispenser roll,
then the end of the tape traces a curve called the involute of the circle. Let's
find the parametric equations for this curve, assuming that the dispensing roll
has constant radius a and is centered at the origin. (As more and more tape is
unwound, the radius of the roll will, of course, decrease. We'll assume that little
enough tape is unwound so that the radius of the roll remains constant.)
Considering Figure 1 .29, we see that the position vector O P of the desired
point P is the vector sum OB + BP. To determine O B and BP, we use the angle
9 between the positive x-axis and OB as our parameter. Since B is a point on the
circle,
OB = a cos# i + a sin9 j.
Unwound
\ tape
B / Involute
\ 0
J (a, 0)
Figure 1 .29 Unwinding tape, as
in Example 8. The point P
describes a curve known as the
involute of the circle.
Figure 1 .30 The vector BP must
make an angle of 9 — n/2 with the
positive x-axis.
To find the vector B P, parallel translate it so that its tail is at the origin. Figure 1 .30
shows that B~P 's length must be a9, the amount of unwound tape, and its direction
must be such that it makes an angle of 0 — n/2 with the positive x-axis. From our
experience with circular geometry and, perhaps, polar coordinates, we see that
BP is described by
iO cos — |) i + ad sin (0 — — ^ j = a6 sin0 i — a9 cosO j.
OP = ~0% + B~^ = a(cos6> +6>sin0)i + a(sin6» - 6>cos6>)j.
Ix = a(cos9 + 0 sin#)
y = a(sin 0 — 9 cos 0)
are the parametric equations of the involute, whose graph is pictured in
Figure 1.31. ♦
BP
Hence,
So
Chapter 1 | Vectors
1.2 Exercises
In Exercises 1-5, write the given vector by using the standard
basis vectors for R2 and R3.
1. (2,4) 2. (9,-6)
4. (-1,2,5) 5. (2,4,0)
3. (3,;r, -7)
In Exercises 6—10, write the given vector without using the
standard basis notation.
6. i + j -3k
7. 9i-2j + V2k
8. -3(2i-7k)
9. jri — j (Consider this to be a vector in R2.)
10. jri — j (Consider this to be a vector in R3.)
11. Letai =(1, 1) and a2 = (1,-1).
(a) Write the vector b = (3, 1) as ciai + C2a2, where
c\ and C2 are appropriate scalars.
(b) Repeat part (a) for the vector b = (3, — 5).
(c) Show that any vector b = {b\, bj) in R2 may be
written in the form c^i + C2&2 for appropriate
choices of the scalars c\, cj, (This shows that ai
and a2 form a basis for R2 that can be used instead
of i and j.)
12. Leta!=(l,0, -l),a2 = (0, 1, 0), and a3 =(1, 1,-1).
(a) Find scalars c\, c%, C3, so as to write the vector
b = (5, 6, —5) as Qai + C2a2 + c^.
(b) Try to repeat part (a) for the vector b = (2, 3, 4).
What happens?
(c) Can the vectors ai, a2, a3 be used as a basis for
R3, instead of i, j, k? Why or why not?
In Exercises 13—18, give a set of parametric equations for the
lines so described.
1 3. The line in R3 through the point (2, — 1 , 5) that is par-
allel to the vector i + 3j — 6k.
14. The line in R3 through the point (12, -2, 0) that is
parallel to the vector 5i — 12j + k.
1 5. The line in R2 through the point (2, — 1 ) that is parallel
to the vector i — 7j.
16. The line in R3 through the points (2,1,2) and
(3,-1,5).
17. The line in R3 through the points (1,4,5) and
(2,4,-1).
18. The line in R2 through the points (8, 5) and (1, 7).
1 9. Write a set of parametric equations for the line in R4
through the point (1, 2, 0, 4) and parallel to the vector
(-2,5,3,7).
20. Write a set of parametric equations for the line in
R5 through the points (9,^,-1,5,2) and (-1,1,
V2,7, 1).
21 . (a) Write a set of parametric equations for the line in
R3 through the point (— 1 , 7, 3) and parallel to the
vector 2i — j + 5k.
(b) Write a set of parametric equations for the line
through the points (5,-3, 4) and (0, 1, 9).
(c) Write different (but equally correct) sets of equa-
tions for parts (a) and (b).
(d) Find the symmetric forms of your answers in
(a)-(c).
22. Give a symmetric form for the line having parametric
equations x = 5 — 2t, y = 3t + 1, z = 6t — 4.
23. Give a symmetric form for the line having parametric
equations x = t + 1, y = 3t — 9, z = 6 — 8f .
24. A certain line in R3 has symmetric form
x - 2 v-3 z+l
5 -2 4
Write a set of parametric equations for this line.
25. Give a set of parametric equations for the line with
symmetric form
x + 5 y-1 z+10
3 7-2
26. Are the two lines with symmetric forms
x - I y + 2 z+l
and
x - 4 y-1 z + 5
10 -5 :
the same? Why or why not?
27. Show that the two sets of equations
z x + 1 y + 6 z + 5
- and = '- =
5 _6 -14 -10
x — 2 y — 1
3 = 7
actually represent the same line in R3 .
28. Determine whether the two lines / 1 and I2 defined by
the sets of parametric equations h: x = 2t — 5, y =
3t + 2, z = 1 - 6t, and l2: x = 1 - 2t, y = 11 - 3f,
z = 6t — 17 are the same. (Hint: First find two points
on l\ and then see if those points lie on I2.)
29. Do the parametric equations l\\ x = 3t + 2, y =
t - 7, z = 5t+l, and l2: x = 6t - 1, y = 2t - 8,
z = 10? — 3 describe the same line? Why or why not?
1.2 | Exercises
30. Do the parametric equations x = 3f3 + 7, y = 2 — f3,
Z = 5/3 + 1 determine a line? Why or why not?
31. Do the parametric equations x = 5t2 — 1, y = 2t2 +
3, z = 1 — t2 determine a line? Explain.
32. A bird is flying along the straight-line path x = 2t + 7,
y = t — 2, z= 1 — 3f, where/ is measured in minutes.
(a) Where is the bird initially (at t = 0)? Where is the
bird 3 minutes later?
(b) Give a vector that is parallel to the bird's path.
(c) When does the bird reach the point ( y , | , — y )?
(d) Does the bird reach (17, 4, -14)?
33. Find where the line x = 3t — 5, y = 2 — t, z = 6t in-
tersects the plane x + 3y — z = 19.
34. Where does the line x = 1 — At, y = t — 3/2, z =
2t + 1 intersect the plane 5x — 2y + z = 1?
35. Find the points of intersection of the line x = 2t — 3,
y = 3t +2, z = 5 — t with each of the coordinate
planes x = 0, y = 0, and z = 0.
36. Show that the line x = 5 — t, y = 2t — 7, z = t — 3 is
contained in the plane having equation 2x — y + 4z = 5.
37. Does the line x = 5 — t, y = 2t — 3, z = It + 1 inter-
sect the plane x — 3y + z = 1? Why?
38. Find where the line having symmetric form
x — 3_y + 2_z
6 ~ 3 ~ 5
intersects the plane with equation 2x — 5v + 3z + 8 = 0.
39. Show that the line with symmetric form
x -3 z+2
^r = y-5 = —
lies entirely in the plane 3x + 3 y + z = 22.
40. Does the line with symmetric form
x+4 _ y-2 _ z-\
3 ~ -1 ~ -9
intersect the plane 2x — 3y + z = 7?
41 . Let a,b,cbe nonzero constants. Show that the line with
parametric equations x = at + a, y = b, z = ct + c
lies on the surface with equation x2/a2 + y2/b2 —
Z2/c2 = 1.
42. Find the point of intersection of the two lines l\ : x =
2t + 3, y = 3t + 3, z = 2t + 1 and l2:x = 15 - It,
y = t-2,z = 3t-l.
43. Do the lines lx:x = 2t+\, y = -3t, z = t-\
and l2:x = 3t + 1, y = t + 5, z = 7 -t intersect?
Explain your answer.
44. (a) Find the distance from the point (—2, 1, 5) to any
point on the line x = 3t — 5, y = 1 — t,z = 4t + 7.
(Your answer should be in terms of the param-
eter t.)
(b) Now find the distance between the point (—2, 1 , 5)
and the line x = 3t - 5, y = 1 - t, z = At + 7.
(The distance between a point and a line is the dis-
tance between the given point and the closest point
on the line.)
45. (a) Describe the curve given parametrically by
x = 2cos3r „ 27r
0 < t < — .
y = 2 sin3r 3
What happens if we allow t to vary between 0
and 27r?
(b) Describe the curve given parametrically by
x = 5cos3f „ 27r
0 < t < — .
y = 5 sin3r 3
(c) Describe the curve given parametrically by
{x = 5 sin3t „ 27r
0 < t < — .
y = 5 cos3f 3
(d) Describe the curve given parametrically by
x = 5 cos 3t „ 27r
0 < t < — .
y = 3 sin3r 3
46. Suppose that a bicycle wheel of radius a rolls along a
fiat surface without slipping. If a reflector is attached
to a spoke of the wheel at a distance b from the center,
the resulting curve traced by the reflector is called a
curtate cycloid. One such cycloid appears in Fig-
ure 1.32, where a = 3 and b = 2.
y
1 1
1
2ji
i
4?r
Figure 1 .32 A curtate cycloid.
Using vector methods or otherwise, find a set of para-
metric equations for the curtate cycloid. Figure 1.33
should help. (Take a low point of the cycloid to lie
1 8 Chapter 1 | Vectors
Figure 1 .34 A prolate cycloid.
Figure 1 .33 The point P traces a
curtate cycloid.
on the v-axis.) There is no theoretical reason that the
cycloid just described cannot have a < b, although in
such case the bicycle-wheel-reflector application is no
longer relevant. (When a < b, the parametrized curve
that results is called a prolate cycloid.) Your paramet-
ric equations should be such that the constants a and b
can be chosen independently of one another. An exam-
ple of a prolate cycloid, with a = 2 and b = 4, is shown
in Figure 1.34. Try to think of a physical situation in
which such a curve would arise.
47. Egbert is unwinding tape from a circular dispenser of
radius a by holding the tape taut and perpendicular to
the dispenser. Find a set of parametric equations for
the path traced by the end of the tape (the point P in
Figure 1 .35) as Egbert unwinds the tape. Use the angle
0 between OP and the positive x -axis for parameter.
Assume that little enough tape is unwound so that the
radius of the dispenser remains constant.
p
/
u
Figure 1.35 Figure for Exercise 47.
1.3 The Dot Product
When we introduced the arithmetic notions of vector addition and scalar mul-
tiplication, you may well have wondered why the product of two vectors was
not defined. You might think that "vector multiplication" should be defined in a
manner analogous to the way we defined vector addition (i.e., by componentwise
multiplication). However, such a definition is not very useful. Instead we shall
define and use two different concepts of a product of two vectors: (1) the Eu-
clidean inner product, or "dot" product, which may be defined for two vectors in
R" (where n is arbitrary) and (2) the "cross" or vector product, which is defined
only for vectors in R3 .
1.3 | The Dot Product 19
The Dot Product of Two Vectors
DEFINITION 3.1 Let a = (a\,a2, a3) and b = (bu b2, b3) be two vectors
in R3. The dot (or inner or scalar) product of a and b, denoted a • b, is
a • b = a\b\ + a2b2 + 03^3 ■
In R2, the analogous definition is
a • b = a\b\ + a2b2,
where a = (a\ , a2) and b = (b\ , b2).
EXAMPLE 1 In R3, we have
(1, -2, 5) • (2, 1, 3) = (1)(2) + (-2X1) + (5X3) = 15.
(3i + 2j - k) • (i - 2k) = (3X1) + (2X0) + (-1X-2) = 5. #
In accordance with its name, the dot — or scalar — product takes two vectors
and produces a single real number (not a vector).
The following facts are consequences of Definition 3.1:
Properties of dot products. If a, b, and c are any vectors in R3 (or R2) and
k e R is any scalar, then
1. a • a > 0, and a • a = 0 if and only if a = 0;
2. a-b = ba;
3. a • (b + c) = a • b + a • c;
4. (jfca) • b = jfc(a • b) = a • (kb).
Proof of Property 1 If a = (a\, a2, a^), then we have
a • a = a\ci\ + a2a2 + aj,a^ = a\ + a\+ a\.
This last expression is evidently nonnegative, since it is a sum of squares of real
numbers. Moreover, such an expression is zero exactly when each of the terms is
zero, that is, if and only if a\ = a2 = = 0. ■
We leave the proofs of properties 2, 3, and 4 as exercises.
Thus far, we have introduced the dot product of two vectors as a purely
algebraic construction. It is the geometric interpretation of the definition that is
really interesting. To establish this interpretation, we begin with the following:
DEFINITION 3.2 If a = (a\,a2, a3), then the length of a (also called the
norm or magnitude), denoted ||a||, is ^/a2x + a2 + a2.
Chapter 1 | Vectors
a
Figure 1.36 The dot
product of a and b is
llall llbll cosd.
a
Figure 1 .37 The vector
triangle used in the proof
of Theorem 3.3.
The motivation for this definition is evident if we draw a as the position vector of
the point (a\ , a%, a$). Then the length of the arrow from the origin to {a\ , 02, 03) is
V(ai - 0)2 + (a2 " 0)2 + (a3 - 0)2,
as given by the distance formula, which is nothing more than an extension of the
Pythagorean theorem in the plane. As we just saw, a ■ a = a\ + a\ + a\, and we
have
||a|| = v/aT a
or, equivalently,
a - a = ||a||2. (1)
Now we're ready to state the main result concerning the geometry of the dot
product. If a and b are two nonzero vectors in R3 (or R2) drawn with their tails at
the same point, let 9, where 0 < 9 < it, be the angle between a and b. If either a
or b is the zero vector, then 9 is indeterminate (i.e., can be any angle).
THEOREM 3.3 If a and b are any two vectors in either R2 or R3, then
a-b= ||a|| ||b|| cos (9.
(See Figure 1.36.)
PROOF If either a or b is the zero vector, say a, then a = (0, 0, 0) and so
a.b = (0)(&i) + (0)(&2) + (0)0>3) = 0.
Also, || a|| = 0 in this case, so the formula in Theorem 3.3 holds. In this case, the
angle 6 is indeterminate.
Now suppose that neither a nor b is the zero vector. Let c = b — a. Then
we may apply the law of cosines to the triangle whose sides are a, b, and c
(Figure 1.37) to obtain
Hell2 = ||a||2 + ||b||2 -2||a|| ||b|| cose.
Thus,
2 1| a|| ||b|| cose = ||a||2 + ||b||2 - ||c||2 = a • a + b • b - c • c, (2)
from equation (1). Now, use the properties of the dot product. Since c = b — a,
c • c = (b - a) • (b — a)
= (b - a) • b - (b - a) • a
= b- b- a- b- b- a + a-a, (3)
by properties 3 and 4 of the dot product. If we use equation (3) to substitute for
c • c in equation (2), then
2 1| a || ||b|| cos6» = a-a + b-b-(b-b-a-b-b-a + a-a)
= a • b + b • a
= 2a-b,
by property 2 of the dot product. By canceling the factor of 2 on both sides, the
desired result is obtained. ■
1.3 | The Dot Product 21
Figure 1 .38 An object
sliding down a ramp.
The force due to gravity
is downward, but the
direction of travel of the
object is inclined 30° to
the horizontal.
Angles Between Vectors
Theorem 3.3 may be used to find the angle between two nonzero vectors a and
b — just solve for 6 in the formula in Theorem 3.3 to obtain
The use of the inverse cosine is unambiguous, since we take 0 < 6 < it when
defining angles between vectors.
EXAMPLE 2 If a = i + j and b = j - k, then formula (4) gives
= cos
-i (i + j)-(j
= cos
1
= COS
-1
1
7T
3'
+ jllllj-k|| (V2-V2)
If a and b are nonzero, then Theorem 3.3 implies
cos 6 = 0 if and only if a • b = 0.
We have cos 6> = Ojust incased = jt/2. (Remember our restriction on 0.) Hence,
it makes sense for us to call a and b perpendicular (or orthogonal) when a • b =
0. If either a or b is the zero vector, then we cannot use formula (4), and the angle
9 is undefined. Nonetheless, since a • b = 0 if a or b is 0, we adopt the standard
convention and say that the zero vector is perpendicular to every vector.
EXAMPLE 3 The vector i + j is orthogonal to the vector i — j + k, since
(i + j) • (i - j + k) = (1X1) + (IX- 1) + (0X1) = 0- ♦
Vector Projections
Suppose that a 2 kg object is sliding down a ramp having a 30° incline with the
horizontal as in Figure 1.38. If we neglect friction, the only force acting on the
object is gravity. What is the component of the gravitational force in the direction
of motion of the object?
To answer questions of this nature, we need to find the projection of one vector
on another. The general idea is as follows: Given two nonzero vectors a and b,
imagine dropping a perpendicular line from the head of b to the line through a.
Then the projection of b onto a, denoted projab, is the vector represented by the
arrow in Figure 1.39.
projab projab
Figure 1 .39 Projection of the vector b onto the vector a.
22 Chapter 1 | Vectors
Given this intuitive understanding of the projection, we find a precise formula
for it. Recall that a vector is determined by magnitude (length) and direction. It
follows by definition that the direction of projab is either the same as that of a,
or opposite to a if the angle 9 between a and b is more than jt/2. Trigonometry
then tells us
, m II Projab ||
cos# = - -.
(The absolute value sign around cos# is needed in case jt/2 < 0 < jt.) Hence,
with a bit of algebra, we have
II -kII mi .I I m ||a|| ||b|||cos0| |a-b|
||proJab|| = ||b|||cos0| = = —
by Theorem 3.3. Thus, we know the magnitude and direction of projab. To obtain
a compact formula for projab, note the following:
PROPOSITION 3.4 Let k be any scalar and a any vector. Then
1. II ka\\ = \k\ ||a||.
2. A unit vector (i.e., a vector of length 1) in the direction of a nonzero vector
a is given by a/||a||.
PROOF Part 1 is left as an exercise. (Write out ka and ||&a|| in terms of compo-
nents.) For part 2, we must check that the length of a/||a|| is 1:
a
1
a
Ilall
Ilall
|a|| = 1,
by part 1 (since l/||a|| is a positive scalar).
Now projab is a vector of length a • b|,
|a-b|\ a
Tal
projab = ±
| a || in the "±a-direction." That is,
_± ||a|| ||b|| |cosfj| a
length of
projab
unit vector
in direction of a
Note that the angle 9 keeps track of the appropriate sign of projab; that is, when
0 < 6 < jt/2, cos 0 is positive and projab points in the direction of a, and when
7r/2 < 8 < Tt, cos 9 is negative and projab points in the direction opposite to that
of a. Thus, we can eliminate both the ± sign and the absolute value, and we find
that
||a|| ||b|| cos<9 a a-b
proi„b = = r a
Ilall ||a|| ||a||2
by Theorem 3.3, so that
by equation (1). Formula (5) is concise and not difficult to remember.
1.3 | The Dot Product
Figure 1 .40 The 2 kg
object sliding down a
ramp in Example 4.
EXAMPLE 4 To answer the question posed at the beginning of this subsection,
we need to calculate projaF, where F is the gravitational force vector and a points
along the ramp as shown in Figure 1 .40. We have a coordinate situation as shown
in Figure 1.41. From trigonometric considerations, we must have a = a\\ + a^j
such that flj = — ||a|| cos 30° and a2 = — ||a|| sin 30°. Since we are really only
interested in the direction of a, there is no loss in assuming that a is a unit vector.
Thus,
a = - cos 30° i- sin 30° j = -^i- \\.
al^b°
F = -mgj = -19.6j
Figure 1 .41 The vectors a and F in Example 4,
realized in a coordinate system.
Taking g = 9.8 m/sec , we have F
implies
-2gj = — 19. 6j. Therefore, formula (5)
. a-F
projaF = | | a =
a • a
(- |j)-(-19.6j) / V3. O
27
« -8.49i - 4.9 j,
and the component of F in this direction is
||projaF|| = ||-8.49i-4.9j|| =9.8N. ♦
Unit vectors — that is, vectors of length 1 — are important in that they capture
the idea of direction (since they all have the same length). Part 2 of Proposition
3.4 shows that every nonzero vector a can have its length adjusted to give a unit
vector u = a/ 1| a|| that points in the same direction as a. This operation is referred
to as normalization of the vector a.
EXAMPLE 5 A fluid is flowing across a plane surface with uniform velocity
vector v. If n is a unit vector perpendicular to the plane surface, let's find (in terms
of v and n) the volume of the fluid that passes through a unit area of the plane in
unit time. (See Figure 1.42.)
24 Chapter 1 | Vectors
Figure 1 .42 Fluid flowing across
a plane surface.
Base has area 1
Figure 1 .43 After one unit of time, the fluid passing
across a square will have filled the box.
First, imagine one unit of time has elapsed. Then over a unit area of the plane
(say over a unit square), the fluid will have filled a "box" as in Figure 1.43. The
box may be represented by a parallelepiped (a three-dimensional analogue of a
parallelogram). The volume we seek is the volume of this parallelepiped and is
Volume = (area of base)(height).
The area of the base is 1 unit by construction. The height is given by ||projnv||.
From formula (5),
/n-v\
Pr°Jnv = ( ) n = (n • v)n,
vn • n/
since n • n = ||n||2 = 1. Hence,
||projnv|| = ||(n-v)n|| = |n-v| ||n|| = |n-v|,
by part 1 of Proposition 3.4. ♦
Vector Proofs
We conclude this section with two illustrations of how wonderfully well vectors
are suited to providing elegant proofs of geometric results.
EXAMPLE 6 In an arbitrary triangle, show that the line segment joining the
midpoints of two sides is parallel to and has half the length of the third side. (See
Figure 1.44.) In other words, if M\ is the midpoint of side AB and M2 is the
midpoint of side AC, we wish to show that M\ M2 is parallel to BC and has half
its length.
A A
Figure 1 .44 In triangle ABC, Figure 1 .45 The vector version
Mi M2 is parallel to B C and has of triangle ABC in Example 6.
half its length.
For a vector proof, we use the diagram in Figure 1 .45, a slightly modified ver-
sion of Figure 1 .44. The midpoint conditions translate to the following statements
about vectors:
AA?i = ±A$, A~\f2 = \At.
1.3 | The Dot Product
Now,
M1M2 = AM2 - ~AM\ = \AC - \lh = \{AC - AB) = \~B~£.
But M\ M2 = \BC is precisely what we wish to prove: To say M\M2 is a scalar
times BC means that the two vectors are parallel. Moreover, from part 1 of
Proposition 3.4,
\\Mjt2\\ = \\\BC\\ = I||fi£||,
so that the length condition also holds. ♦
EXAMPLE 7 Show that every angle inscribed in a semicircle is a right angle,
as suggested by Figure 1 .46.
Figure 1.46 Every angle Figure 1.47 a and b are "radius
inscribed in a semicircle is a vectors."
right angle.
To prove this remark, we'll make use of Figure 1 .47, where a and b are "radius
vectors" with tails at the center of the circle. We need only show that a — b (a
vector along one ray of the angle in question) is perpendicular to — a — b (a vector
along the other ray). In other words, we wish to show that
(a - b) • (-a - b) = 0.
We have
(a - b) • (-a - b) = (-l)(a - b) • (a + b),
by property 4 of dot products,
= (-l)((a-b)-a + (a-b)-b)
= (-l)(a a-b a + a b-b b)
= (-l)(||a||2- ||b||2),
by properties 2 and 4,
= 0,
since both a and b are radius vectors (and therefore have the same length, namely,
the radius of the circle). ♦
Vector proofs as in Examples 6 and 7 are elegant and sometimes allow you to
write shorter and more direct proofs than those from your high school geometry
days.
26 Chapter 1 [ Vectors
1.3 Exercises
Compute a • b, ||a||, ||b|| for the vectors listed in Exercises 1—6.
1.
a =
(l,5),b = (-2,3)
2.
a =
(4,-l),b = (±,2)
3.
a =
(-1,0, 7), b = (2, 4, -6)
4.
a =
(2, 1, 0), b = (1, -2, 3)
5.
a =
4i - 3j + k, b = i + j + k
6.
a =
i + 2j-k,b = -3j + 2k
In Exercises 7-11, find the angle between each of the pairs of
vectors.
7.
a =
V3i + j,b = — a/3 i + j
8.
a
(-1,2), b = (3, 1)
9.
a =
i + j,b = i + j + k
10.
a =
i + j - k, b = -i + 2j + 2k
11.
a =
(l,-2,3),b = (3,-6, -5)
In Exercises 12—16, calculate projab.
12. a = i + j, b = 2i + 3j - k
13. a = (i + j)/V2, b = 2i + 3j-k
14. a = 5k, b = i-j + 2k
15. a= -3k,b = i-j + 2k
16. a = i + j + 2k, b = 2i - 4j + k
1 7. Give a unit vector that points in the same direction as
the vector 2i — j + k.
1 8. Give a unit vector that points in the direction opposite
to the vector — i + 2k.
1 9. Give a vector of length 3 that points in the same direc-
tion as the vector i + j — k.
20. Find three nonparallel vectors that are perpendicular
to i - j + k.
21. Is it ever the case that projab = projba? If so, under
what conditions?
22. Prove properties 2, 3, and 4 of dot products.
23. Prove part 1 of Proposition 3.4.
24. Suppose that a force F = i — 2j is acting on an object
moving parallel to the vector a = 4i + j. Decompose F
into a sum of vectors Fi and F2, where F; points along
the direction of motion and F2 is perpendicular to the
direction of motion. (Hint: A diagram may help.)
25. In physics, when a constant force acts on an object
as the object is displaced, the work done by the force
is the product of the length of the displacement and
the component of the force in the direction of the dis-
placement. Figure 1 .48 depicts an object acted upon by
a constant force F, which displaces it from the point P
to the point Q. Let 6 denote the angle between F and
the direction of displacement.
(a) Show that the work done by F is determined by the
formula F- Pg.
(b) Find the work done by the (constant) force F =
i + 5j + 2k in moving a particle from the point
(1, -1, 1) to the point (2, 0, -1).
F
6L
>Q
Component of F in direction
of displacement
Figure 1.48 A constant force F
displaces the object from P to Q. (See
Exercise 25.)
26. A refrigerator is dragged 12 ft across a smooth floor
using a rope and 60 lb of force directed along the rope.
How much work is done if the rope makes a 20° angle
with the horizontal?
27. How much work is done in pushing a handtruck loaded
with 500 lb of bananas 40 ft up a ramp inclined 30°
from horizontal?
Let a be a nonzero vector in R3. The direction cosines of a are
the three numbers cos a, cos ft, cos y determined by the angles
a, fi, y between a and, respectively, the positive x-, y-, and
Z-axes. In Exercises 28 and 29, find the direction cosines of the
given vectors.
28. a
29. a
i + 2j-k
3i + 4k
30. If a = flii + 02) + 03k, give expressions for the direc-
tion cosines of a in terms of the components of a.
31. Let A, P, and C denote the vertices of a triangle. Let
0 < r < 1 . If Pi is the point on AS located r times the
distance from A to 5 and Pi is the point on AC located
r times the distance from A to C, use vectors to show
that Pi P2 is parallel to PC and has r times the length
of PC. (This result generalizes that of Example 6 of
this section.)
1.4 | The Cross Product
32. Let A, B, C, and D be four points in R3 such that no
three of them lie on a line. Then ABCD is a quadri-
lateral, though not necessarily one that lies in a plane.
Denote the midpoints of the four sides of ABCD by
Mi, M%, M3, and M4. Use vectors to show that, amaz-
ingly, M1M2M2M4 is always a parallelogram.
33. Use vectors to show that the diagonals of a parallel-
ogram have the same length if and only if the paral-
lelogram is a rectangle. (Hint: Let a and b be vectors
along two sides of the parallelogram. Express vectors
running along the diagonals in terms of a and b. See
Figure 1.49.)
a
Figure 1.49 Diagram for Exercise 33.
34. Using vectors, prove that the diagonals of a parallelo-
gram are perpendicular if and only if the parallelogram
is a rhombus. (Note: A rhombus is a parallelogram
whose four sides all have the same length.)
35. This problem concerns three circles of equal radius r
that intersect in a single point O. (See Figure 1.50.)
(a) Let W\, W2, and W3 denote the centers of the
three circles and let w,- = OWi for i = 1, 2, 3.
Similarly, let A, B, and C denote the remaining
intersection points of the circles and set a = OA,
b = OB, and c = OC. By numbering the centers
of the circles appropriately, write a, b, and c in
terms of wi , W2, and W3 .
(b) Show that A, B, and C lie on a circle of the
same radius r as the three given circles. (Hint:
The center of the circle is at the point P, where
OP = Wl + W2 + W3 .)
Figure 1.50 Two examples
of three circles of equal radius
intersecting in a single
point O. (See Exercise 35.)
(c) Show that O is the orthocenter of triangle ABC.
(The orthocenter of a triangle is the common in-
tersection point of the altitudes perpendicular to
the edges.)
36. (a) Show that the vectors [|b[|a+ [|a||b and ||b||a —
|| a || b are orthogonal.
(b) Show that ||b||a + ||a||b bisects the angle between
a and b.
1 .4 The Cross Product
The cross product of two vectors in R3 is an "honest" product in the sense that it
takes two vectors and produces a third one. However, the cross product possesses
some curious properties (not the least of which is that it cannot be defined for
vectors in R2 without first embedding them in R3 in some way) making it less
"natural" than may at first seem to be the case.
When we defined the concepts of vector addition, scalar multiplication, and
the dot product, we did so algebraically (i.e., by a formula in the vector compo-
nents) and then saw what these definitions meant geometrically. In contrast, we
will define the cross product first geometrically, and then deduce an algebraic for-
mula for it. This technique is more convenient, since the coordinate formulation
28 Chapter 1 | Vectors
a
Figure 1 .51 The area of this
parallelogram is ||a|| ||b|| sin#.
Figure 1 .53 i x j = k.
Figure 1 .54 A mnemonic for
finding the cross product of the
unit basis vectors.
is fairly complicated (although we will find a way to organize it so as to make it
easier to remember).
The Cross Product of Two Vectors in R3
DEFINITION 4.1 Let a and b be two vectors in R3 (not R2). The cross
product (or vector product) of a and b, denoted a x b, is the vector whose
length and direction are given as follows:
• The length of a x b is the area of the parallelogram spanned by a and b
or is zero if either a is parallel to b or if a or b is 0. Alternatively, the
following formula holds:
||axb|| = ||a|| ||b|| sin 6,
where 0 is the angle between a and b. (See Figure 1.51.)
• The direction of a x b is such that a x b is perpendicular to both a and
b (when both a and b are nonzero) and is taken so that the ordered triple
(a, b, a x b) is a right-handed set of vectors, as shown in Figure 1.52.
(If either a or b is 0, or if a is parallel to b, then a x b = 0 from the
aforementioned length condition.)
By saying that (a, b, a x b) is right-handed we mean that if you let the fingers
of your right hand curl from a toward b, then your thumb will point in the
direction of a x b.
EXAMPLE 1 Let's compute the cross product of the standard basis vectors
for R3. First consider i x j as shown in Figure 1.53. The vectors i and j deter-
mine a square of unit area. Thus, ||i x j|| = 1. Any vector perpendicular to both
i and j must be perpendicular to the plane in which i and j lie. Hence, i x j
must point in the direction of ±k. The "right-hand rule" implies that i x j must
point in the positive k direction. Since ||k|| = 1, we conclude that i x j = k. The
same argument establishes that j x k = i and k x i = j. To remember these ba-
sic equations, you can draw i, j, and k in a circle, as in Figure 1.54. Then the
relations
ixj = k, jxk = i, kxi = j (1)
may be read from the circle by beginning at any vector and then proceeding
clockwise. ♦
Properties of the Cross Product; Coordinate Formula
Example 1 demonstrates that the calculation of cross products from the geometric
definition is not entirely routine. What we really need is a coordinate formula,
analogous to that for the dot product or for vector projections, which is not difficult
to obtain.
1.4 I The Cross Product 29
From our Definition 4.1, it is possible to establish the following:
Properties of the Cross Product. Let a, b, and c be any three vectors in R3
and let k e R be any scalar. Then
1. axb = — bxa (anticommutativity);
2. ax(b + c) = axb + axc (distributivity);
3. (a + b)xc = axc + bxc (distributivity);
4. k(a x b) = (fca) x b = a x (kb).
We provide proofs of these properties at the end of the section, although you
might give some thought now as to why they hold. It's worth remarking that these
properties are entirely reasonable, ones that we would certainly want a product to
have. However, you should be clear about the fact that the cross product fails to
satisfy other properties that you might also consider to be eminently reasonable.
In particular, since property 1 holds, we see that axb/bxain general (i.e.,
the cross product is not commutative). Consequently, be very careful about the
order in which you write cross products. Another property that the cross product
does not possess is associativity. That is,
a x (b x c) / (a x b) x c,
in general. For example, let a = b = i and c = j. Then
i x (i x j) = i x k = -k x i = -j,
from properties 1 and 4, but (ixi)xj = 0xj = 0^— j. (The equation
i x i = 0 holds because i is, of course, parallel to i.) Make sure that you do
not try to use an associative law when working problems.
We now have the tools for producing a coordinate formula for the cross
product. Let a = a\i + a2\ + a^k and b = b\\ + b2\ + b^k. Then
a x b = (a\i + a2\ + a^k) x (b\\ + b2\ + b^k)
= (aii + a2\ + 03k) x bii + (flii + a2\ + a3k) x b2j
+ (aii + a2) + 03k) x Z?3k,
by property 2,
= aib\\ x i + a2b\j x i + c^ik x i + a\b2\ x j + a2b2\ x j
+ a^b2k x j + aib^i x k + a2b^i x k + a^b^k x k,
by properties 3 and 4. These nine terms may look rather formidable at first, but
we can simplify by means of the formulas in (1), anticommutativity, and the fact
that c x c = 0 for any vector c e R3. (Why?) Thus,
a x b = —a2b\k + «3i>ij + a\b2k — a^b2i — a\bi] + #2^31
= (a2b3 - a3b2)\ + (a3bi - + {axb2 - a2bx)k. (2)
EXAMPLE 2 Formula (2) gives
(i + 3j - 2k) x (2i + 2k) = (3 • 2 - (-2) • 0)i + (-2 • 2 - 1 • 2)j
+ (1 • 0 - 3 • 2)k
= 6i - 6j - 6k. ♦
Chapter 1 | Vectors
Formula (2) is more complicated than the corresponding formulas for all the
other arithmetic operations of vectors that we've seen. Moreover, it is a rather dif-
ficult formula to remember. Fortunately, there is a more elegant way to understand
formula (2). We explore this reformulation next.
Matrices and Determinants: A First Introduction —
A matrix is a rectangular array of numbers. Examples of matrices are
1 2 3
4 5 6
"l
3
2
7
and
0
0
1 0 0 0"
0 10 0
0 0 10
0 0 0 1
If a matrix has m rows and n columns, we call it "m x n" (read "m by w").
Thus, the three matrices just mentioned are, respectively, 2 x 3,3 x 2, and 4 x 4.
To some extent, matrices behave algebraically like vectors. We discuss some
elementary matrix algebra in § 1 .6. For now, we are mainly interested in the notion
of a determinant, which is a real number associated toann x « (square) matrix.
(There is no such thing as the determinant of a nonsquare matrix.) In fact, for the
purposes of understanding the cross product, we need only study 2x2 and 3x3
determinants.
DEFINITION 4.2 Let A be a 2 x 2 or 3 x 3 matrix. Then the determinant
of A, denoted det A or | A|, is the real number computed from the individual
entries of A as follows:
• 2 x 2 case
If A =
• 3 x 3 case
, then |A| =
a
b c
d
e f
, then
J
h i
a b
c
\A
d e
f
g h
i
= ad — be.
= aei +bfg + cdh — ceg — afh — bdi
e f
h i
d f
g i
+ c
d e
g h
in terms of 2 x 2 determinants.
Perhaps the easiest way to remember and compute 2x2 and 3x3 determi-
nants (but not higher-order determinants) is by means of a "diagonal approach."
We write (or imagine) diagonal lines running through the matrix entries. The
determinant is the sum of the products of the entries that lie on the same diagonal,
1.4 | The Cross Product
where negative signs are inserted in front of the products arising from diagonals
going from lower left to upper right:
•2x2 case
A =
and
\A\ = ad — be.
3x3 case (we need to repeat the first two columns for the method to work)
a b c
d e f
_ g h i
Write
A =
Then
\A\ = aei + bfg + cdh — gee — hfa — idb.
Important Warning This mnemonic device does not generalize beyond
3x3 determinants.
We now state the connection between determinants and cross products.
Key Fact. If a = a\\ + d2] + a^k and b = b\i + bi\ + ^k, then
j +
a x b
a2 a3
b2 h
bx b3
b\ b2
k =
i j k
a,\ a2 «3
b\ b2 b}
(3)
The determinants arise from nothing more than rewriting formula (2). Note
that the 3 x 3 determinant in formula (3) needs to be interpreted by using the 2 x 2
determinants that appear in formula (3). (The 3x3 determinant is sometimes
referred to as a "symbolic determinant.")
EXAMPLE 3
(3i + 2j-k)x(i-j + k)
2 -1
3 -1
j +
3 2
1 1
i —
1 1
1 -1
= i - 4j - 5k
32 Chapter 1 | Vectors
We may also calculate the 3 x 3 determinant as
= 2i - j - 3k - 2k - i - 3j = i - 4j - 5k.
Areas and Volumes
Cross products are used readily to calculate areas and volumes of certain objects.
We illustrate the ideas involved with the next two examples.
EXAMPLE 4 Let's use vectors to calculate the area of the triangle whose ver-
tices are A(3, 1), B(2, —1), and C(0, 2) as shown in Figure 1.55.
B
Figure 1 .55
Triangle ABC in
Example 4.
Figure 1 .56 Any triangle may be
considered to be half of a
parallelogram.
The trick is to recognize that any triangle can be thought of as half of a
parallelogram (see Figure 1.56) and that the area of a parallelogram is obtained
from a cross product. In other words, A B x A C is a vector whose length measures
the area of the parallelogram determined by and AC, and so
Area of AABC = UlA^ x ACjl.
To use the cross product, we must consider Tb and AC to be vectors in R3 .
is straightforward: We simply take the k-components to be zero. Thus,
This
AB =
2j =
2j - Ok,
and
Therefore,
AC = — 3i + j = -3i + j + 0k.
Tb
x AC =
j k
-2 0
1 0
= -7k.
Hence,
Area of AABC = i|| - 7k|| = \. ♦
1.4 | The Cross Product
There is nothing sacred about using A as the common vertex. We could just
as easily have used B or C, as shown in Figure 1.57. Then
Area of AABC = \\\b\ x B~C\\ = ~||(i-l-2j) x (-2i + 3j)|| = ±||7k|| = \.
EXAMPLE 5 Find a formula for the volume of the parallelepiped determined
by the vectors a, b, and c. (See Figure 1.58.)
a x b
B
Figure 1.57 The area of AABC
is 7/2.
|c|| |cos 6\
a
Figure 1 .58 The parallelepiped determined by a, b,
and c.
As explained in §1.3, the volume of a parallelepiped is equal to the product
of the area of the base and the height. In Figure 1.58, the base is the parallelogram
determined by a and b. Hence, its area is ||a x b||. The vector a x b is perpendi-
cular to this parallelogram; the height of the parallelepiped is || c || | cos 6 1 , where 9
is the angle between a x b and c. (The absolute value is needed in case 9 > tt/2.)
Therefore,
Volume of parallelepiped = (area of base)(height)
= ||a x b|| ||c|| |cos0|
= |(axb)-c|.
(The appearance of the cos 9 term should alert you to the fact that dot products
are lurking somewhere.)
For example, the parallelepiped determined by the vectors
a = i + 5 j , b = — 4i + 2j, and c = i + j + 6k
has volume equal to
|((i + 5j)x(-4i + 2j)).(i + j + 6k)| = |22k-(i + j + 6k)|
= |22(6)|
= 132. ♦
The real number (a x b) • c appearing in Example 5 is known as the triple
scalar product of the vectors a, b, and c. Since | (a x b) • c| represents the volume
of the parallelepiped determined by a, b, and c, it follows immediately that
|(a x b)-c| = |(b x c)-a| = |(c x a)-b|.
34 Chapter 1 | Vectors
Figure 1 .59 Turning a bolt with a
wrench. The torque on the bolt is
the vector r x F.
Figure 1.60 A
potato spinning
about an axis.
Figure 1 .61 The
angular velocity
vector a>.
In fact, if you are careful with the right-hand rule, you can convince yourself that
the absolute value signs are not needed; that is,
(a x b) • c = (b x c) • a = (c x a) • b. (4)
This is a nice example of how the geometric significance of a quantity can provide
an extremely brief proof of an algebraic property the quantity must satisfy. (Try
proving it by writing out the expressions in terms of components to appreciate
the value of geometric insight.)
We leave it to you to check the following beautiful (and convenient) formula
for calculating triple scalar products:
(a x b) • c ■
ai
(22 ^3
C2 C3
where a = a\\ + ay + «3k, b = b\\ + + bj,k, and c = c\i + C2] + c^k.
Torque
Suppose you use a wrench to turn a bolt. What happens is the following: You
apply some force to the end of the wrench handle farthest from the bolt and that
causes the bolt to move in a direction perpendicular to the plane determined by
the handle and the direction of your force (assuming such a plane exists). To
measure exactly how much the bolt moves, we need the notion of torque (or
twisting force).
In particular, letting F denote the force you apply to the wrench, we have
Amount of torque = (length of wrench)(component of F _L wrench).
Let r be the vector from the center of the bolt head to the end of the wrench
handle. Then
where 9 is the ang
torque is ||r x F||,
as the direction in
bolt). Hence, it is
torque vector T is
Note that if F
fact that if you try
Amount of torque = ||r|| ||F|| sin#,
;le between r and F. (See Figure 1.59.) That is, the amount of
and it is easy to check that the direction of r x F is the same
which the bolt moves (assuming a right-handed thread on the
quite natural to define the torque vector T to be r x F. The
a concise way to capture the physics of this situation,
is parallel to r, then T = 0. This corresponds correctly to the
to push or pull the wrench, the bolt does not turn.
Rotation of a Rigid Body
Spin an object (a rigid body) about an axis as shown in Figure 1.60. What is the
relation between the (linear) velocity of a point of the object and the rotational
velocity? Vectors provide a good answer.
First we need to define a vector co, the angular velocity vector of the rotation.
This vector points along the axis of rotation, and its direction is determined by
the right-hand rule. The magnitude of (o is the angular speed (measured in radians
per unit time) at which the object spins. Assume that the angular speed is constant
in this discussion. Next, fix a point O (the origin) on the axis of rotation, and let
r(r) = OP~ be the position vector of a point P of the body, measured as a function
of time, as in Figure 1.61. The velocity v of P is defined by
Ar
v = hm — ,
Ar^O At
1.4 | The Cross Product 35
t(0\
Figure 1 .62 A spinning rigid
body.
where Ar = r(t + At) — r(t) (i.e., the vector change in position between times t
and t + At). Our goal is to relate v and co.
As the body rotates, the point P (at the tip of the vector r) moves in a circle
whose plane is perpendicular to co. (See Figure 1 .62, which depicts the motion
of such a point of the body.) The radius of this circle is ||r(?)|| sin#, where 0 is
the angle between co and r. Both ||r(/)|| and 6 must be constant for this rotation.
(The direction of r(t) may change with t, however.) If At « 0, then ||Ar|| is
approximately the length of the circular arc swept by P between / and t + At.
That is,
|| Ar || «s (radius of circle)(angle swept through by P)
= (||r|| sin0)(A0)
from the preceding remarks. Thus,
Ar
At
sinO
A<p
Now, let At -> 0. Then Ar/Ar -> v and Acp/At
angular velocity vector co, and we have
co r sine
(o x r
\co\\ by definition of the
(5)
Figure 1 .63 A carousel wheel.
It's not difficult to see intuitively that v must be perpendicular to both co and r.
A moment's thought about the right-hand rule should enable you to establish the
vector equation
v = co x r.
(6)
If we apply formula (5) to a bicycle wheel, it tells us that the speed of a point
on the edge of the wheel is equal to the product of the radius of the wheel and
the angular speed (9 is n/2 in this case). Hence, if the rate of rotation is kept
constant, a point on the rim of a large wheel goes faster than a point on the rim
of a small one. In the case of a carousel wheel, this result tells you to sit on an
outside horse if you want a more exciting ride. (See Figure 1.63.)
Summary of Products Involving Vectors
Following is a collection of some basic information concerning scalar multipli-
cation of vectors, the dot product, and the cross product:
Scalar Multiplication: ka
Result is a vector in the direction of a.
Magnitude is ||&a|| = \k\ ||a||.
Zero if k = 0 or a = 0.
Commutative: ka = ak.
Associative: k(la) = (kl)a.
Distributive: £(a + b) = ka + kb; (k + l)a = ka + la.
Chapter 1 | Vectors
Dot Product: a • b
Result is a scalar
Magnitude is a • b = Hall 1 hi cos (9*
G is the an)?1e between a and b
rVTapnitnde is maximized ifa II Yt
Zero if a _L b, a = 0, or b = 0.
Commutative: a • b = b • a.
Associativity is irrelevant, since (a ■
b) • c doesn't make sense.
Distributive: a-(b + c) = a- b + a
• c.
Ifa = b, then a- a = ||a||2.
Cross Product: a x b
Result is a vector perpendicular to both a and b.
Magnitude is ||a x b|| = ||a|| ||b|| sin#; 0 is the angle between a and b.
Magnitude is maximized if a _L b.
Zero ifa || b, a = 0, orb = 0.
Anticommutative: axb = — bxa.
Not associative: In general, a x (b x c) ^ (a x b) x c.
Distributive: ax(b + c) = axb + axc and
(a + b)xc = axc + bxc.
Ifa _Lb, then ||axb|| = ||a|| ||b||.
Addendum: Proofs of Cross Product Properties
Proof of Property 1 To prove the anticommutativity property, we use the right-
hand rule. Since
||axb|| = ||a|| ||b|| sin0,
we obviously have that ||a x b|| = ||b x a||. Therefore, we need only understand
the relation between the direction of a x b and that of b x a. To determine the
direction of a x b, imagine curling the fingers of your right hand from a toward b.
Then your thumb points in the direction of a x b. If instead you curl your fingers
from b toward a, then your thumb will point in the opposite direction. This is the
direction of b x a, so we conclude that a x b = — b x a. (See Figure 1.64.) ■
ax b
Figure 1 .64 The right-hand rule shows why a x b = — b x a.
1.4 | The Cross Product 37
Proof of Property 2 First, note the following general fact:
PROPOSITION 4.3 Let a and b be vectors in R3 . If a • x = b • x for all vectors x
in R3, then a = b.
To establish Proposition 4.3, write a as aii + ci2] + 03k and b as b\i + +
b^k and set x in turn equal to i, j, and k. Proposition 4.3 is valid for vectors in R2
as well as R3 .
To prove the distributive law for cross products (property 2), we show that,
for any x e R3 ,
(a x (b + c)) • x = (a x b + a x c) • x.
By Proposition 4.3, property 2 follows.
From the equations in (4),
(a x (b + c)) • x = (x x a) • (b + c)
= (x x a) • b + (x x a) • c,
from the distributive law for dot products,
= (a x b) • x + (a x c) • x
= (a x b + a x c) • x,
again using (4) and the distributive law for dot products. ■
Proof of Property 3 Property 3 follows from properties 1 and 2. We leave the
details as an exercise. ■
Proof of Property 4 The second equality in property 4 follows from the first
equality and property 1 :
k(a x b) = —k(b x a) by property 1
= —(kb) x a by the first equality of property 4
= a x (kb) by property 1 .
Hence, we need only prove the first equality.
If either a or b is the zero vector or if a is parallel to b, then the first equality
clearly holds. Otherwise, we divide into three cases: (1) k = 0, (2) k > 0, and
(3) k < 0. If k = 0, then both ka and k(a x b) are equal to the zero vector and
the desired result holds. If k > 0, the direction of (ka) x b is the same as a x b,
which is also the same as k(a x b). Moreover, the angle between ka and b is the
same as between a and b. Calling this angle 0, we check that
||(*a)xb|| = ||*a|| ||b|| sine
= &||a|| ||b|| sin# by part 1 of Proposition 3.4
= k || a x b || by Definition 4. 1
= \\k(a x b)|| by part 1 of Proposition 3.4.
We conclude (ka) x b = k(a x b) in this case.
38 Chapter 1 | Vectors
a
fca, k < 0
fca, k > 0
Figure 1 .65 If the angle between
a and b is 0, then the angle
between £a and b is either 0 (if
k > 0) or?r -0 (if£ < 0).
If k < 0, then the direction of (ka) x b is the same as that of (—a) x b, which
is seen to be the same as that of —(a x b) and thus the same as that of k(a x b).
The angle between ka and b is therefore n — 9, where 9 is the angle between a
and b. (See Figure 1.65.) Thus,
||(*a) x b|| = ||&a|| ||b|| sin(7r - 9) = \k\ ||a|| ||b|| sin 9 = ||*(a x b)||.
So, again, it follows that (ka) x b = k(a x b). ■
1.4 Exercises
Evaluate the determinants in Exercises 1—4.
1.
3.
2.
4.
In Exercises 5—7, calculate the indicated cross products, using
both formulas (2) and (3).
5. (1, 3, -2) x (-1,5, 7)
6. (3i-2j + k)x(i + j + k)
7. (i + j)x(-3i + 2j)
8. Prove property 3 of cross products, using properties 1
and 2.
9. If a x b = 3i - 7j - 2k, what is (a + b) x (a - b)?
1 0. Calculate the area of the parallelogram having vertices
(1,1), (3, 2), (1,3), and (-1,2).
1 1 . Calculate the area of the parallelogram having vertices
(1,2, 3), (4, -2, 1), (-3, 1, 0), and (0, -3, -2).
1 2. Find a unit vector that is perpendicular to both 2i +
j — 3k and i + k.
13. If (a x b) • c = 0, what can you say about the geomet-
ric relation between a, b, and c?
Compute the area of the triangles described in Exercises
14-17.
1 4. The triangle determined by the vectors a = i + j and
b = 2i-j
15. The triangle determined by the vectors a = i — 2j +
6k and b = 4i + 3j - k
16. The triangle having vertices (1,1), (—1,2), and
(-2,-1)
17. The triangle having vertices (1,0, 1), (0,2,3), and
(-1,5,-2)
1 8. Find the volume of the parallelepiped determined by
a = 3i - j, b = -2i + k, and c = i - 2j + 4k.
1 9. What is the volume of the parallelepiped with vertices
(3,0,-1), (4,2,-1), (-1,1,0), (3,1,5), (0,3,0),
(4,3,5), (-1,2, 6), and (0,4,6)?
20. Verify that (a x b) • c
a i
a2 a-}
b2 b3
C2 c3
21 . Show that (a x b) • c = a • (b x c) using Exercise 20.
22. Use geometry to show that |(axb)-c| =
|b-(axc)|.
23. (a) Show that the area of the triangle with vertices
P\(xi,y\), P2(x2, yi), and P3(x3, y3) is given by
the absolute value of the expression
1 1 1
x\ x2 JC3
yi yi yi
(b) Use part (a) to find the area of the triangle with
vertices (1,2), (2, 3), and (-4, -4).
24. Suppose that a, b, and c are noncoplanar vectors in R3 ,
so that they determine a tetrahedron as in Figure 1 .66.
Figure 1.66 The tetrahedron of
Exercise 24.
Give a formula for the surface area of the tetrahedron
in terms of a, b, and c. (Note: More than one formula
is possible.)
1.4 | Exercises 39
25. Suppose that you are given nonzero vectors a, b, and c
in R3 . Use dot and cross products to give expressions for
vectors satisfying the following geometric descriptions:
(a) A vector orthogonal to a and b
(b) A vector of length 2 orthogonal to a and b
(c) The vector projection of b onto a
(d) A vector with the length of b and the direction of a
(e) A vector orthogonal to a and b x c
(f) A vector in the plane determined by a and b and
perpendicular to c.
26. Suppose a, b, c, and d are vectors in R3 . Indicate which
of the following expressions are vectors, which are
scalars, and which are nonsense (i.e., neither a vector
nor a scalar).
(a) (a x b) x c
(c) (a • b) x (c • d)
(e) (a • b) x (c x d)
(g) (a x b) • (c x d)
(b) (a-b)-c
(d) (a x b) • c
(f) ax[(b-c)d]
(h) (a • b)c - (a x b)
Exercises 27—32 concern several identities for vectors a, b, c,
and d in R3. Each of them can be verified by hand by writing
the vectors in terms of their components and by using formula
(2) for the cross product and Definition 3.1 for the dot product.
However, this is quite tedious to do. Instead, use a computer
algebra system to define the vectors a, b, C, and d in general
and to verify the identities.
^27. (a x b) x c = (a • c)b - (b • c)a
^ 28. a • (b x c) = b • (c x a) = c • (a x b)
= -a • (c x b) = -c • (b x a)
= -b • (a x c)
^ 29. (a x b) • (c x d) = (a • c)(b • d) - (a • d)(b • c)
a • c a • d
b-c b-d
^ 30. (a x b) x c + (b x c) x a + (c x a) x b = 0 (this is
known as the Jacobi identity).
^ 31. (a x b) x (c x d) = [a • (c x d)]b - [b • (c x d)]a
^ 32. (a x b) • (b x c) x (c x a) = [a • (b x c)]2
33. Establish the identity
(a x b) • (c x d) = (a • c)(b • d) - (a • d)(b • c)
of Exercise 29 without resorting to a computer algebra
system by using the results of Exercises 27 and 28.
34. Egbert applies a 20 lb force at the edge of a 4 ft
wide door that is half-open in order to close it. (See
Figure 1.67.) Assume that the direction of force is per-
pendicular to the plane of the doorway. What is the
torque about the hinge on the door?
35. Gertrude is changing a flat tire with a tire iron. The tire
iron is positioned on one of the bolts of the wheel so
Figure 1.67 Figure for Exercise 34.
40 lb
Figure 1.68 The configuration for
Exercise 35.
that it makes an angle of 30° with the horizontal. (See
Figure 1.68.) Gertrude exerts 40 lb of force straight
down to turn the bolt.
(a) If the length of the arm of the wrench is 1 ft, how
much torque does Gertrude impart to the bolt?
(b) What if she has a second tire iron whose length is
18 in?
36. Egbert is trying to open a jar of grape jelly. The ra-
dius of the lid of the jar is 2 in. If Egbert imparts 15 lb
of force tangent to the edge of the lid to open the jar,
how many ft-lb, and in what direction, is the resulting
torque?
37. A 50 lb child is sitting on one end of a seesaw, 3 ft
from the center fulcrum. (See Figure 1 .69.) When she is
1.5 ft
Figure 1.69 The seesaw of Exercise 37.
40 Chapter 1 | Vectors
1 .5 ft above the horizontal position, what is the amount
of torque she exerts on the seesaw?
38. For this problem, note that the radius of the earth is
approximately 3960 miles.
(a) Suppose that you are standing at 45° north latitude.
Given that the earth spins about its axis, how fast
are you moving?
(b) How fast would you be traveling if, instead, you
were standing at a point on the equator?
39. Archie, the cockroach, and Annie, the ant, are on an
LP record. Archie is at the edge of the record (ap-
proximately 6 in from the center) and Annie is 2 in
closer to the center of the record. How much faster is
Archie traveling than Annie? (Note: A record playing
on a turntable spins at a rate of 33 j revolutions per
minute.)
40. A top is spinning with a constant angular speed of 12
radians/sec. Suppose that the top spins about its axis
of symmetry and we orient things so that this axis is
the z-axis and the top spins counterclockwise about it.
(a) If, at a certain instant, a point P in the top has
coordinates (2,-1, 3), what is the velocity of the
point at that instant?
(b) What are the (approximate) coordinates of P one
second later?
41. There is a difficulty involved with our definition of
the angular velocity vector go, namely, that we cannot
properly consider this vector to be "free" in the sense
of being able to parallel translate it at will. Consider
the rotations of a rigid body about each of two parallel
axes. Then the corresponding angular velocity vectors
go1 and <w2 are parallel. Explain, perhaps with a fig-
ure, that even if G0l and G02 are equal as "free vectors,"
the corresponding rotational motions that result must
be different. (Therefore, when considering more than
one angular velocity, we should always assume that the
axes of rotation pass through a common point.)
Figure 1 .70 The plane in R3
through the point Pq and
perpendicular to the vector n.
1 .5 Equations for Planes; Distance Problems
In this section, we use vectors to derive analytic descriptions of planes in R3. We
also show how to solve a variety of distance problems involving "fiat objects"
(i.e., points, lines, and planes).
Coordinate Equations of Planes
A plane IT in R3 is determined uniquely by the following geometric information:
a particular point Pq(xo, vo, zo) in the plane and a particular vector n = Ai +
B\ + Ck that is normal (perpendicular) to the plane. In other words, n is the
set of all points P(x, y, z) in space such that PqP is perpendicular to n. (See
Figure 1.70.) This means that n is defined by the vector equation
n- Kf> = 0.
(1)
Since PqP = (x — xo)i + (y — yo)j + (z — zo)k, equation (1) may be rewritten
as
(Ai + fij + Ck) • ((x - x0)i + (y - y0)j + (z - z0)k) = 0
or
A(x - x0) + B(y - y0) + C(z - z0) = 0.
(2)
This is equivalent to
Ax + By + Cz = D,
where D = Axq + Byo + Czo-
1.5 | Equations for Planes; Distance Problems 41
EXAMPLE 1 Theplanethroughthepoint(3,2, 1) with normal vector 2i — j + 4k
has equation
(2i - j + 4k) • ((x - 3)i + (y - 2)j + (z - l)k) = 0
<=► 2(x - 3) - (y - 2) + 4(z - 1) = 0
2x - y + 4z = 8. #
Not only does a plane in R3 have an equation of the form given by equation
(2), but, conversely, any equation of this form must describe a plane. Moreover,
it is easy to read off the components of a vector normal to the plane from such an
equation: They are just the coefficients of x, y, and z.
EXAMPLE 2 Given the plane with equation Ix + 2y — 3z = 1, find a normal
vector to the plane and identify three points that lie on that plane.
A possible normal vector is n = 7i + 2j — 3k. However, any nonzero scalar
multiple of n will do just as well. Algebraically, the effect of using a scalar multiple
of n as normal is to multiply equation (2) by such a scalar.
Finding three points in the plane is not difficult. First, let y = z = 0 in the
defining equation and solve for x:
Ix + 2-0-3-0=1 <^=> 7x = l x=f
Thus (j, 0, 0) is a point on the plane. Next, let x = z = 0 and solve for y:
7-0 + 2^-3-0=1 y=\.
So (0, j, 0) is another point on the plane. Finally, let x = y = 0 and solve for z.
You should find that (0,0,— j) lies on the plane. ♦
EXAMPLE 3 Put coordinate axes on R3 so that the z-axis points vertically.
Then a plane in R3 is vertical if its normal vector n is horizontal (i.e., if n is
parallel to the xy-plane). This means that n has no k-component, so n can be
written in the form Ai + Bj. It follows from equation (2) that a vertical plane has
an equation of the form
A(x - x0) + B(y - yo) = 0.
Hence, a nonvertical plane has an equation of the form
A(x - x0) + B(y - y0) + C(z - z0) = 0,
where C / 0. ♦
EXAMPLE 4 From high school geometry, you may recall that a plane is
determined by three (noncollinear) points. Let's find an equation of the plane
that contains the points P0(l, 2, 0), Pi(3, 1, 2), and P2(0, 1, 1).
There are two ways to solve this problem. The first approach is algebraic
and rather uninspired. From the aforementioned remarks, any plane must have
an equation of the form Ax + By + Cz = D for suitable constants A, B, C, and
D. Thus, we need only to substitute the coordinates of Po, Pi, and P2 into this
equation and solve for A, B, C, and D. We have that
• substitution of Po gives A + 2B = D;
• substitution of Pi gives 3 A + B + 2C = D; and
• substitution of P2 gives B + C = D.
42 Chapter 1 | Vectors
Figure 1.71 The plane
determined by the points Pq, P\,
and P2 in Example 4.
Hence, we must solve a system of three equations in four unknowns:
A + 2B = D
3 A + B + 2C = D .
B + C = D
(3)
In general, such a system has either no solution or else infinitely many solutions.
We must be in the latter case, since we know that the three points Pq, P\, and P2
lie on some plane (i.e., that some set of constants A, B, C, and D must exist).
Furthermore, the existence of infinitely many solutions corresponds to the fact
that any particular equation for a plane may be multiplied by a nonzero constant
without altering the plane defined. In other words, we can choose a value for one
of A, B, C, or D, and then the other values will be determined. So let's multiply
the first equation given in (3) by 3, and subtract it from the second equation. We
obtain
A+ 2B = D
-5B +2C = -2D .
B + C = D
(4)
Now, multiply the third equation in (4) by 5 and add it to the second:
A + 2B = D
1C = 3D . (5)
B + C = D
Multiply the third equation appearing in (5) by 2 and subtract it from the first:
A -2C = -D
1C= 3D . (6)
B + C = D
By adding appropriate multiples of the second equation to both the first and third
equations of (6), we find that
1C =
3D
(7)
B
= %D
Thus, if in (7) we take D = —1 (for example), then A = 1, B = —4, C = —3,
and the equation of the desired plane is
x - 4y - 3z = -7.
The second method of solution is cleaner and more geometric. The idea is
to make use of equation (1). Therefore, we need to know the coordinates of a
particular point on the plane (no problem — we are given three such points) and
a vector n normal to the plane. The vectors P0P1 and P0P2 both lie in the plane.
(See Figure 1.71.) In particular, the normal vector n must be perpendicular to
them both. Consequently, the cross product provides just what we need. That is,
we may take
n = P0Pi x P0P2 = (2i - j + 2k) x (-i - j + k)
= i - 4j - 3k.
1.5 | Equations for Planes; Distance Problems 43
2y + z = 4
Figure 1 .72 The line of
intersection of the planes
x — 2y + z = 4 and
2x + y + 3z = —7 in Example 5.
If we take Po(l, 2, 0) to be the particular point in equation (1), we find that the
equation we desire is
(i-4j-3k)-((x- l)i + (v-2)j + zk) = 0
or
(x - l) - My - 2) - 3z = 0.
This is the same equation as the one given by the first method. ♦
EXAMPLE 5 Consider the two planes having equations x — 2 y + z = 4 and
2x + y + 3z = — 7. We determine a set of parametric equations for their line of
intersection. (See Figure 1.72.) We use Proposition 2.1. Thus, we need to find a
point on the line and a vector parallel to the line. To find the point on the line,
we note that the coordinates (x, y, z) of any such point must satisfy the system of
simultaneous equations given by the two planes
x-2y + z= 4
2x + y + 3z = -7
(8)
From the equations given in (8), it is not too difficult to produce a single
solution (x, y, z). For example, if we let z = 0 in (8), we obtain the simpler
system
(9)
-2,
The solution to the system of equations (9) is readily calculated to be x =
y = —3. Thus, (—2, —3, 0) are the coordinates of a point on the line.
To find a vector parallel to the line of intersection, note that such a vector
must be perpendicular to the two normal vectors to the planes. The normal vectors
to the planes are i — 2j + k and 2i + j + 3k. Therefore, a vector parallel to the
line of intersection is given by
(i - 2j + k) x (2i + j + 3k) = -7i - j + 5k.
Hence, Proposition 2.1 implies that a vector parametric equation for the line is
r(r) = (-2i - 3j) + f(-7i - j + 5k),
and a standard set of parametric equations is
x = -7f - 2
y = -t-3 .
z = 5t ♦
Parametric Equations of Planes
Another way to describe a plane in R3 is by a set of parametric equations. First,
suppose that a = (ct\, a-i, a^) and b = (b\, b2, b^) are two nonzero, nonparallel
vectors in R3. Then a and b determine a plane in R3 that passes through the
origin. (See Figure 1.73.) To find the coordinates of a point P(x, y, z) in this
plane, draw a parallelogram whose sides are parallel to a and b and that has two
opposite vertices at the origin and at P, as shown in Figure 1 .74. Then there must
exist scalars s and t so that the position vector of P is ssi + tb. The plane
44 Chapter 1 | Vectors
z
Figure 1 .73 The plane through
the origin determined by the
vectors a and b.
z
y
Figure 1 .74 For the point P in
the plane shown, OP = .?a + fb
for appropriate scalars s and t.
Figure 1 .75 The plane passing
through Pq(c\ , C2, C3) and parallel
to a and b.
may be described as
{x e R3 | x = sa + fb; s, t e R} .
Now, suppose that we seek to describe a general plane n (i.e., one that does
not necessarily pass through the origin). Let
c = (ci, c2, c3) = OPq
denote the position vector of a particular point P0 in n and let a and b be two
(nonzero, nonparallel) vectors that determine the plane through the origin parallel
to n. By parallel translating a and b so that their tails are at the head of c (as in
Figure 1.75), we adapt the preceding discussion to see that the position vector of
any point P(x, y, z) in n may be described as
= sa + tb + c.
To summarize, we have shown the following:
PROPOSITION 5.1 A vector parametric equation for the plane n containing
the point Pq(c\, c2, c?,) (whose position vector is OPq = c) and parallel to the
nonzero, nonparallel vectors a and b is
x(s, t) = sa+tb + c. (10)
By taking components in formula (10), we readily obtain a set of parametric
equations for n :
x = sa\ + tb\ + C\
y = sa2 + tb2 + c2 ■ (11)
Z = scii + tb-i + c3
Compare formula (10) with that of equation (1) in Proposition 2.1. We need
to use two parameters s and t to describe a plane (instead of a single parameter
t that appears in the vector parametric equation for a line) because a plane is a
two-dimensional object.
1.5 | Equations for Planes; Distance Problems 45
EXAMPLE 6 We find a set of parametric equations for the plane that passes
through the point (1,0,-1) and is parallel to the vectors 3i — k and 2i + 5j + 2k.
From formula (10), any point on the plane is specified by
x(s, t) = s(3i - k) + r(2i + 5j + 2k) + (i
= (3s + 2t + l)i + 5t\ + (2/ - s -
The individual parametric equation may be read off as
x = 3s + 2t + 1
-k)
l)k.
y = 5t
z = 2t
1
Figure 1.76 A general
configuration for finding the
distance between a point and a
line, using vector projections.
Figure 1 .77 Another general
configuration for finding the
distance between a point and a
line.
Distance Problems
Cross products and vector projections provide convenient ways to understand a
range of distance problems involving lines and planes: Several examples follow.
What is important about these examples are the vector techniques for solving
geometric problems that they exhibit, not the general formulas that may be derived
from them.
EXAMPLE 7 (Distance between a point and a line) We find the distance
between the point P0(2, 1 , 3) and the line 1(0 = t(- 1 , 1 , -2) + (2, 3 , -2) in two
ways.
Method 1. From the vector parametric equations for the given line, we read
off a point B on the line — namely, (2, 3, —2) — and a vector a parallel to the
line — namely, a = (— 1, 1,-2). Using Figure 1.76, the length of the vector
BPq — projaSPo provides the desired distance between P0 and the line. Thus,
we calculate that
B~?q = {2, 1,3) -(2, 3,-2)
= (0, -2, 5);
projaBP0 =
a- BP,
o
a • a
(-1, 1,-2). (0,-2, 5)
(-1,1,-2)
" V("l,l,-2)-(-l,l,-2X
= (2, -2, 4).
The desired distance is
||BP0-projafiP0|| = 11(0, -2, 5) -(2, -2, 4)|| = ||(-2, 0,
Method 2. In this case, we use a little trigonometry. If 0 denotes the angle
between the vectors a and BPq as in Figure 1.77, then
D
sin 6* =
= V~5.
\\BP0\
where D denotes the distance between Pq and the line. Hence,
|| a || || Mil sine HaxBPol
D= \\BP0\\ sin<9 =
a
46 Chapter 1 | Vectors
Therefore, we calculate
a x B~fi) =
so that the distance sought is
D =
i j k
-1 1 -2
0-2 5
|i + 5j + 2k||
-i + j-2k|| 76
which agrees with the answer obtained by Method 1 .
= i + 5j + 2k,
Figure 1 .78 The general
configuration for finding the
distance D between two parallel
planes.
/CJL — c
Bo
Figure 1.79 Configuration for
determining the distance between
two skew lines in Example 9.
EXAMPLE 8 (Distance between parallel planes) The planes
111: 2x -2y+z = 5 and n2: 2x - 2y + z = 20
are parallel. (Why?) We see how to compute the distance between them.
Using Figure 1.78 as a guide, we see that the desired distance D is given by
||projn/,iiD2||, where Pi is a point on Oi, P2 is a point on n2, and n is a vector
normal to both planes.
First, the vector n that is normal to both planes may be read directly from
the equation for either n 1 or Tl2 as n = 2i — 2j + k. It is not hard to find a point
Pi on TI 1 : the point Pi(0, 0, 5) will do. Similarly, take P2(0, 0, 20) for a point on
n2. Then
P,P2 = (0,0, 15),
and calculate
projnP1P2 =
n-PiP2
n • n
n =
(2,-2, 1). (0,0, 15)
(2.
= -¥(2,
-2,1) -(2,
v-, "2, 1)
= -§(2,-2,1).
Hence, the distance D that we seek is
D
-2, 1)
(2,-2,1)
llprojnPi/>2
\V9
EXAMPLE 9 (Distance between two skew lines) Find the distance between
the two skew lines
li(0 = *(2. 1,3) + (0, 5,-1) and l2(r) = f(l, -1, 0) + (-1, 2, 0).
(Two lines in R3 are said to be skew if they are neither intersecting nor parallel.
It follows that the lines must lie in parallel planes and that the distance between
the lines is equal to the distance between the planes.)
To solve this problem, we need to find ||projn5i B2 1| , the length of the projec-
tion of the vector between a point on each line onto a vector n that is perpendicular
to both lines, hence, also perpendicular to the parallel planes that contain the lines.
(See Figure 1.79.)
From the vector parametric equations for the lines, we read that the point
Bi(0, 5, —1) is on the first line and 52(— 1, 2, 0) is on the second. Hence,
B,B2 = (-1, 2, 0) - (0, 5, -1) = (-1, -3, 1).
1.5 | Exercises 47
For a vector n that is perpendicular to both lines, we may use n = ai x a2, where
ai = (2, 1 , 3) is a vector parallel to the first line and a2 = ( 1 , — 1 , 0) is parallel to
the second. (We may read these vectors from the parametric equations.) Thus,
n = ai x a2
and so,
projn£i£2
n- BiB2
n • n
J
= 3i + 3j - 3k,
(-1,-3, l)-(3,3,-3)
(3,3, -3) -(3, 3, -3)
(3,3,-3)
(3,3,
= -f(l,l,
-3)
!)•
The desired distance is ||projnJSi Z?2 II = |V3.
1.5 Exercises
1 . Calculate an equation for the plane containing the point
(3,-1,2) and perpendicular to i — j + 2k.
2. Find an equation for the plane containing the point
(9, 5, —1) and perpendicular to i — 2k.
3. Find an equation for the plane containing the points
(3,-1,2), (2, 0,5), and (1, -2,4).
4. Find an equation for the plane containing the points
(A, 0, 0), (0, B, 0), and (0, 0, C). Assume that at least
two of A, B, and C are nonzero.
5. Give an equation for the plane that is parallel to the
plane 5x — 4y + z = 1 and that passes through the
point (2, —1, —2).
6. Give an equation for the plane parallel to the plane 2x —
3y + z = 5 that passes through the point (— 1 , 1,2).
7. Find an equation for the plane parallel to the plane x —
y + Iz = 10 that passes through the point (—2, 0, 1).
8. Give an equation for the plane parallel to the plane
2x + 2y + z = 5 and that contains the line with para-
metric equations x = 2 — t, y = 2t + 1, z = 3 — 2t.
9. Explain why there is no plane parallel to the plane
5x — 3 y + 2z = 10 that contains the line with para-
metric equations x = t + 4, y = 3t — 2, z = 5 — 2t .
1 0. Find an equation for the plane that contains the line x =
2t - 1, y = 3t + 4, z = 1 - t and the point (2, 5, 0).
11. Find an equation for the plane that is perpendicular
to the line x = It - 5, y = 7 - It, z = 8 - t and that
passes through the point (1, —1,2).
12. Find an equation for the plane that contains the two
lines l\.x = t + 2, y = 3t — 5,z = 5t + 1 andh'.x =
5 - t, y = 3t - 10, z = 9 - 2t.
1 3. Give a set of parametric equations for the line of inter-
section of the planes x + 2y — 3z = 5 and 5x + 5y —
z = 1.
1 4. Give a set of parametric equations for the line through
(5, 0, 6) that is perpendicular to the plane 2x — 3y +
5z = -1.
1 5. Find a value for A so that the planes Sx — 6y + 9 Az =
6 and Ax + y + 2z = 3 are parallel.
16. Find values for A so that the planes Ax — y + z = 1
and 3 Ax ; + Ay — 2z = 5 are perpendicular.
Give a set of parametric equations for each of the planes de-
scribed in Exercises 1 7—22.
17. The plane that passes through the point (—1, 2, 7) and
is parallel to the vectors 2i — 3j + k and i — 5k
18. The plane that passes through the point (2,9,-4)
and is parallel to the vectors — 8i + 2j + 5k and
3i - 4j - 2k
19. The plane that contains the lines h: x = 2t + 5, y =
-3t - 6,z = 4f + 10 and l2: x = 5t - 1, y = lOt +
3,z = lt-2
20. The plane that passes through the three points (0, 2, 1),
(7, -1,5), and (-1,3,0)
21. The plane that contains the line /: x = 3t — 5, y =
10 - 3t, z = 2t + 9 and the point (-2, 4, 7)
48 Chapter 1 | Vectors
22. The plane determined by the equation 2x — 3y +
5z = 30
23. Find a single equation of the form Ax + By + Cz = D
that describes the plane given parametrically as x =
3s - t + 2, y = 4s + t, z = s + 5t + 3. (Hint: Begin
by writing the parametric equations in vector form and
then find a vector normal to the plane.)
24. Find the distance between the point (1 , —2, 3) and the
line I: x = 2t - 5, y = 3 - t, z = 4.
25. Find the distance between the point (2, —1) and the
line/:jc = 3f + 7, y = 5t - 3.
26. Find the distance between the point (—11, 10, 20) and
the line /: x = 5 - t , y = 3, z = It + 8.
27. Determine the distance between the two lines li(f) =
f(8, -1,0) + (-1,3, 5) and I2(f) = f(0, 3, 1) +
(0, 3, 4).
28. Compute the distance between the two lines
h(0 = (t ~ 7)i +(5f + l)j + (3 -2f)k and l2(r) =
4ri + (2-0j + (8f+l)k.
29. (a) Find the distance between the two lines li(f) =
f(3, 1,2) + (4, 0,2) and I2(f) = t(\, 2, 3) +
(2,1,3).
(b) What does your answer in part (a) tell you about
the relative positions of the lines?
30. (a) The lines h(r) = t(l, -1, 5) + (2, 0, -4) and
l2(r) = ?(l,-l,5) + (l,3,-5) are parallel. Ex-
plain why the method of Example 9 cannot be used
to calculate the distance between the lines.
(b) Find another way to calculate the distance. (Hint:
Try using some calculus.)
31. Find the distance between the two planes given by the
equations x — 3 y + 2z = 1 and x — 3y + 2z = 8.
32. Calculate the distance between the two planes
5x - 2y + 2z = 12 and - 10.v + 4y - 4z = 8.
33. Show that the distance d between the two parallel
planes determined by the equations Ax + By + Cz =
Di and Ax + By + Cz = D2 is
d= \Di-D2\
Va2 + b2 + c2'
34. Two planes are given parametrically by the vector
equations
Xl(s, f) = (-3, 4, -9) + s(9, -5, 9) + f(3, -2, 3)
x2(s, t) = (5, 0, 3) + s(-9, 2, -9) + r(-4, 7, -4).
(a) Give a convincing explanation for why these
planes are parallel.
(b) Find the distance between the planes.
35. Write equations for the planes that are parallel to
x + 3y — 5z = 2 and lie three units from it.
36. Suppose that li(r) = fa + bi and l2(f) = ra + b2 are
parallel lines in either R2 or R3 . Show that the distance
D between them is given by
D= ||ax(b2-bi)||
11*11
(Hint: Consider Example 7.)
37. Let n be the plane in R3 with normal vector n that
passes through the point A with position vector a. If b
is the position vector of a point B in R3, show that the
distance D between B and n is given by
|n-(b-a)|
D = -.
1 1 n 1 1
38. Show that the distance D between parallel planes with
normal vector n is given by
|n-(x2 -xQ|
1 1 n 1 1
where X] is the position vector of a point on one of the
planes, and x2 is the position vector of a point on the
other plane.
39. Suppose that li (f) = t&i + bi andl2(f) = fa2 + b2 are
skew lines in R3 . Use the geometric reasoning of Ex-
ample 9 to show that the distance D between these lines
is given by
_ |(at x a2)-(b2 -bQ|
|| ai x a2||
1 .6 Some n-dimensional Geometry
Vectors in R"
The algebraic idea of a vector in R2 or R3 is defined in § 1 . 1 , in which we asked
you to consider what would be involved in generalizing the operations of vector
addition, scalar multiplication, etc., to n-dimensional vectors, where n can be
arbitrary. We explore some of the details of such a generalization next.
1.6 | Some n-dimensional Geometry 49
DEFINITION 6.1 A vector in R" is an ordered rc-tuple of real numbers. We
use a = (til, a2, • • ■ , an) as our standard notation for a vector in R".
EXAMPLE 1 The 5-tuple (2, 4, 6, 8, 10) is a vector in R5. The (n + l)-tuple
(2«, 2rc — 2, 2« — 4, . . . , 2, 0) is a vector in R"+1 , where « is arbitrary. ♦
Exactly as is the case in R2 or R3 , we call two vectors a = (a\ , a2 , . . . , an) and
b = (pi, b2 Z?„) equal just incases, = bt for/ = 1,2,..., n. Vector addition
and scalar multiplication are defined in complete analogy with Definitions 1 .3 and
1.4: If a = (a\, a2, . . . , an) and b = (b\ , b2, . . . , b„) are two vectors in R" and
k e R is any scalar, then
a + b = (ai + b\, a2 + b2 an + fc„)
and
k& = (ka\, ka2, . . . , ^an).
The properties of vector addition and scalar multiplication given in §1.1 hold
(with proofs that are no different from those in the two- and three-dimensional
cases). Similarly, the dot product of two vectors in R" is readily defined:
a • b = fli&i + a2b2 + • • • + a„bn.
The dot product properties given in §1.3 continue to hold in n dimensions; we
leave it to you to check that this is so.
What we cannot do in dimensions larger than three is to develop a pictorial
representation for vectors as arrows. Nonetheless, the power of our algebra and
analogy does allow us to define a number of geometric ideas. We define the length
of a vector in a e R" by using the dot product:
||a|| = Va^a.
The distance between two vectors a and b in R" is
Distance between a and b = ||a — b||.
We can even define the angle between two nonzero vectors by using a generalized
version of equation (4) of § 1 .3 :
Here a, be R" and 9 is taken so that 0 < 6 < it. (Note: At this point in our
discussion, it is not clear that we have
a-b
-1 < < 1,
" Ila|| ||b|| "
which is a necessary condition if our definition of the angle 9 is to make sense.
Fortunately, the Cauchy-Schwarz inequality — formula (1) that follows — takes
care of this issue.) Thus, even though we are not able to draw pictures of vectors
in R", we can nonetheless talk about what it means to say that two vectors are
perpendicular or parallel, or how far apart two vectors may be. (Be careful about
this business. We are defining notions of length, distance, and angle entirely in
Chapter 1 | Vectors
terms of the dot product. Results like Theorem 3.3 have no meaning in R", since
the ideas of angles between vectors and dot products are not independent.)
There is no simple generalization of the cross product. However, see Exer-
cises 39^12 at the end of this section for the best we can do by way of analogy.
We can create a standard basis of vectors in R" that generalize the i, j,
k-basis in R3 . Let
d =(1,0, 0.....0),
e2=(0, 1,0,..., 0),
e„ = (0, 0, . . . , 0, 1).
Then it is not difficult to see (check for yourself) that
a = (ai, a2, . . . , a„) = a\&\ + a2e2 H h a„e„
Here are two famous (and often handy) inequalities:
PROOF If n = 2 or 3, this result is virtually immediate in view of Theorem 3.3.
However, in dimensions larger than three, we do not have independent notions of
inner products and angles, so a different proof is required.
First note that the inequality holds if either a or b is 0. So assume that a and
b are nonzero. Then we may define the projection of b onto a just as in §1.3:
/a-b\
projab = I — I a = ka.
Here k is, of course, the scalar a • b/a • a. Let c = b — ka (so that b = ka + c).
Then we have a • c = 0, since
a • c
a • a
= a
(b-
- jfca)
= a
b-
ka • a
/a-b
= a
b-
\a • a
= a
b-
a-b
= 0.
We leave it to you to check that the "Pythagorean theorem" holds, namely, that
the following equation is true:
|b||2 = £2||a||2 + ||c||
Multiply this equation by || a
2
\2 -
a • a. We obtain
a
b||2 =
|a||2£2||a||2 + ||a||2||c||2
= a
a-b
a • a
|a||2 + ||a||2||c||2
1.6 | Some n-dimensional Geometry
= (»-»)(^) (a.a)+||a||2||c||2
= (a.b)2 + ||a||2||c||2.
Now, the quantity ||a||2||c||2 is nonnegative. Hence,
||a||2||b||2>(a.b)2.
Taking square roots in this last inequality yields the result desired. ■
The geometric motivation for this proof of the Cauchy-Schwarz inequality
comes from Figure 1 .80. 1
PROOF Strategic use of the Cauchy-Schwarz inequality yields
||a + b||2 = (a + b)-(a + b)
= a- a + 2a-b + b- b
< a-a + 2||a|| ||b|| +b-b by(l)
= ||a||2 + 2||a|| ||b|| + ||b||2
= (l|a|| + ||b||)2.
Thus, the result desired holds by taking square roots, since the quantities on both
sides of the inequality are nonnegative. ■
In two or three dimensions the triangle inequality has the following obvious
proof from which the inequality gets its name: Since ||a||, ||b||, and ||a + b|| can
be viewed as the lengths of the sides of a triangle, inequality (2) says nothing
more than that the sum of the lengths of two sides of a triangle must be at least
as large as the length of the third side, as demonstrated by Figure 1.81.
Matrices
We had a brief glance at matrices and determinants in § 1 .4 in connection with the
computation of cross products. Now it's time for another look.
A matrix is defined in § 1 .4 as a rectangular array of numbers. To extend our
discussion, we need a good notation for matrices and their individual entries. We
used the uppercase Latin alphabet to denote entire matrices and will continue to
do so. We shall also adopt the standard convention and use the lowercase Latin
alphabet and two sets of indices (one set for rows, the other for columns) to
identify matrix entries. Thus, the general m x n matrix can be written as
A =
a\\ an ■•■ din
C?21 an ■ ■ ■ ain
- "ml @m2 ' ' ' @mn
= (shorthand) (fly).
1 See J. W. Cannon, Amer. Math. Monthly 96 (1989), no. 7, 630-631.
Chapter 1 | Vectors
The first index always will represent the row position and the second index, the
column position.
Vectors in R" can also be thought of as matrices. We shall have occasion to
write the vector a = (a\, a%, . . . , an) either as a row vector (a 1 x n matrix),
a = [ a\ ai ■ ■ ■ an ] ,
or, more typically, as a column vector (an nxl matrix),
a\
a2
a =
1- "71 — I
We did not use double indices since there is only a single row or column present.
It will be clear from context (or else indicated explicitly) in which form a vector
a will be viewed. An m x n matrix A can be thought of as a "vector of vectors"
in two ways: (1) as m row vectors in R",
[ an an ■ ■ ■ a\n ]
[ <221 a22 ■ ■ ■ a2n ]
A
[ am\ am2
or (2) as n column vectors in R"! ,
nn 1
an
an
a\n
a2\
a2n
- am\ -
- am2 _
- amn _
A =
We now define the basic matrix operations. Matrix addition and scalar mul-
tiplication are really no different from the corresponding operations on vectors
(and moreover, they satisfy essentially the same properties).
DEFINITION 6.2 (Matrix Addition) Let A and B be two m x n matrices.
Then their matrix sum A + B is the m x n matrix obtained by adding cor-
responding entries. That is, the entry in the ith row and 7 th column of A + B
is atj + btj, where a,7 and are the ijth entries of A and B, respectively.
EXAMPLE 2 If
A =
then
However, if B
and B
A + B =
2
10
the same dimensions as A.
, then A + B is not defined, since B does not have
1.6 | Some n-dimensional Geometry
Properties of matrix addition. For all m x n matrices A, B, and C we have
1. A + B = B + A (commutativity);
2. A + (B + C) = (A + B) + C (associativity);
3. An m x n matrix O (the zero matrix) with the property that A + O = A
for all m x n matrices A.
DEFINITION 6.3 (Scalar Multiplication) If A is an m x n matrix and
k e R is any scalar, then the product kA of the scalar k and the matrix A is
obtained by multiplying every entry in A by k. That is, the i/th entry of kA
is kciij (where ay is the iyth entry of A).
EXAMPLE 3 IfA =
1
4
, then 3A =
3 6 c
12 15 If
Properties of scalar multiplication. If A and B are any m x n matrices
and k and / are any scalars, then
1. (k + l)A = kA + lA (distributivity);
2. k(A + B) = kA + kB (distributivity);
3. k(lA) = (kl)A = l(kA).
We leave it to you to supply proofs of these addition and scalar multiplication
properties if you wish.
Just as defining products of vectors needed to be "unexpected" in order to be
useful, so it is with defining products of matrices. To a degree, matrix multipli-
cation is a generalization of the dot product of two vectors.
DEFINITION 6.4 (Matrix Multiplication) Let A be an m x n matrix
and B an n x p matrix. Then the matrix product A B is the m x p matrix
whose i/th entry is the dot product of the j'th row of A and the y'th column
of B (considered as vectors in R"). That is, the ijth entry of
au ayi ■ ■ ■ a\n
[an ai2 ■ ■ ■ ain]
-bn ...
b2\
~b,jl
bij
. . . bip
bip
.bn\
- bnj _
Kp.
is
a
i\b\j + ai2b2j H h a
'n
n
bnj = (more compactly) gaby ■
k=l
54 Chapter 1 | Vectors
EXAMPLE 4 If
A =
1 2 3
4 5 6
and B =
0 1
7 0
2 4
then the (2, 1) entry of AB is the dot product of the second row of A and the first
column of B:
(2, 1) entry = [ 4 5 6 ]
= (4)(0) + (5)(7) + (6)(2) = 47.
The full product A B is the 2x2 matrix
20 13
47 28
On the other hand B A is the 3x3 matrix
4 5 6
7 14 21
18 24 30
Order matters in matrix multiplication. To multiply two matrices we must
have
Number of columns of left matrix = number of rows of right matrix.
In Example 4, the products AB and BA are matrices of different dimensions;
hence, they could not possibly be the same. A worse situation occurs when the
matrix product is defined in one order and not the other. For example, if A is 2 x 3
and £ is 3 x 3, then A B is defined (and is a 2 x 3 matrix), but B A is not. However,
even if both products AB and BA are defined and of the same dimensions (as is
the case if A and B are both n x n, for example), it is in general still true that
AB + BA.
Despite this negative news, matrix multiplication does behave well in a number
of respects, as the following results indicate:
Properties of matrix multiplication. Suppose A, B, and C are matrices
of appropriate dimensions (meaning that the expressions that follow are all
defined) and that k is a scalar. Then
1. A(BC) = (AB)C;
2. k(AB) = (kA)B = A(kB);
3. A(B + C) = AB + AC;
4. (A + B)C = AC + BC.
The proofs of these properties involve little more than Definition 6.4, although
the notation can become somewhat involved as in the proof of property 1 .
One simple operation on matrices that has no analogue in the real number
system is the transpose. The transpose of an m x n matrix A is the n x m matrix
1.6 | Some n-dimensional Geometry
AT obtained by writing the rows of A as columns. For example, if
then AT =
A =
1
4
More abstractly, the ijth entry of AT is a,-,-, the jith entry of A.
The transpose operation turns row vectors into column vectors and vice versa
We also have the following results:
(ATf
A,
(A B)1 = BTAT,
for any matrix A. (3)
where A is m x n and B is n x p. (4)
The transpose will largely function as a notational convenience for us. For
example, consider a, b £ R" to be column vectors. Then the dot product a • b can
be written in matrix form as
■ b = a\b\ + a2b2 + • • • + a„bn = [ a\
C'2
In ]
b2
= arb.
EXAMPLE 5 Matrix multiplication is defined the way it is so that, roughly
speaking, working with vectors or quantities involving several variables can be
made to look as much as possible like working with a single variable. This idea
will become clearer throughout the text, but we can provide an important example
now. A linear function in a single variable is a function of the form f(x) = ax
where a is a constant. The natural generalization of this to higher dimensions
is a linear mapping F: R" — > R", F(x) = Ax, where A is a (constant) m x n
matrix and x £ R". More explicitly, F is a function that takes a vector in R"
(written as a column vector) and returns a vector in R"! (also written as a column).
That is,
F(x) = Ax =
an
an ■
X\
Cl2l
a22 ■
_ am\
@m2
&mn _
The function F has the properties that F(x + y) = F(x) + F(y) for all x, y e R"
and F(kx) = k¥(x) for all x e R", i £ R. These properties are also satisfied by
f(x) = ax, of course. Perhaps more important, however, is the fact that linear
mappings behave nicely with respect to composition. Suppose F is as just defined
and G: R"! — > Rp is another linear mapping defined by G(x) = Bx, where B is a
p x m matrix. Then there is a composite function GoF: R" — > Rp defined by
G o F(x) = G(F(x)) = G(Ax) = fi(Ax) = (BA)x
by the associativity property of matrix multiplication. Note that BA is defined
and is a p x n matrix. Hence, we see that the composition of two linear mappings
is again a linear mapping. Part of the reason we defined matrix multiplication the
way we did is so that this is the case. ♦
EXAMPLE 6 We saw that by interpreting equation (1) in §1.2 inn dimensions,
we obtain parametric equations of a line in R" . Equation (2) of § 1 .5, the equation
56 Chapter 1 | Vectors
for a plane in R3 through a given point (xo, yo, zo) with given normal vector
n = Ai + B] + Ck, can also be generalized to n dimensions:
Ax(xx - bi) + A2(x2 — b2) H h An(x„ - bn) = 0.
If we let A = (Ai, A2, . . . , A„), b = (b\, &2> • • • > b„) ("constant" vectors), and
x = (jci, x%, ■ ■ ■ , x„) (a "variable" vector), then the aforementioned equation can
be rewritten as
A • (x - b) = 0
or, considering A, b, and x as n x 1 matrices, as
Ar(x - b) = 0.
This is the equation for a hyperplane in R" through the point b with normal
vector A. The points x that satisfy this equation fill out an (n — l)-dimensional
subset of R". ♦
At this point, it is easy to think that matrix arithmetic and the vector geometry
of R", although elegant, are so abstract and formal as to be of little practical use.
However, the next example, from the field of economics,2 shows that this is not
the case.
EXAMPLE 7 Suppose that we have n commodities. If the price per unit of the
ith commodity is pi, then the cost of purchasing x, (> 0) units of commodity
i is piXj. If p = (pi, .. . , p„) is the price vector of all the commodities and
x = (jti, . . . , xn) is the commodity bundle vector, then
p • x = pxx\ + p2x2 H V p»x„
represents the total cost of the commodity bundle.
Now suppose that we have an exchange economy, so that we may buy and
sell items. If you have an endowment vector w = (w\ ,...,«>„), where to,- is the
amount of commodity / that you can sell (trade), then, with prices given by the
price vector p, you can afford any commodity bundle x where
p • x < p • w.
We may rewrite this last equation as
p • (x - w) < 0.
In other words, you can afford any commodity bundle x in the budget set
{x | p • (x — w) < 0}. The equation p • (x — w) = 0 defines a budget hyperplane
passing through w with normal vector p. ♦
Determinants
We have already defined determinants of 2 x 2 and 3x3 matrices. (See §1.4.)
Now we define the determinant of any n x n (square) matrix in terms of determi-
nantsof(n — 1) x (n — 1) matrices. By "iterating the definition," we can calculate
any determinant.
2 See D. Saari, "Mathematical complexity of simple economics," Notices of the American Mathematical
Society 42 (1995), no. 2, 222-230.
1.6 | Some n-dimensional Geometry
DEFINITION 6.5 Let A = (ay) be an n x n matrix. The determinant of
A is the real number given by
l)1+2«i2|A12| + --- + (-l)1+Bai„|A1„|,
1) submatrix of A obtained by deleting the
\A\={-X)l+lan\An
where Ay is the (n — 1) x («
rth row and jth column of A.
EXAMPLE 8 IfA =
According to Definition 6.5,
det
1
2
1
3
-2
1
0
5
4
2
-1
0
3
-2
1
1
, then
+-
i—
—i—
-2
0
5
4
>
-1
0
3
>
1
1
2
0
5
4
-1
0
3
1
1
1
2
1
3 "
-2
1
0
5
4
2
-1
0
3
-2
1
1
1
0
5
(-l)1+1(l)det
2
-1
0
-2
1
1
-2
0
5
+ (-
l)1+2(2)det
4
-1
0
3
1
1
" -2
1
5
+ (-
l)1+3(l)det
4
2
0
3
-2
1
" -2
1
0
+ (-
l)1+4(3)det
4
2
-1
3
-2
1
(l)(l)(-l) + (-l)(2)(37) + (l)(l)(-78)
+ (-l)(3)(-7)
-132. ♦
The determinant of the submatrix A/j of A is called the ijth minor of A, and
the quantity (— 1),+; |A,7 | is called the ijth cofactor. Definition 6.5 is known as
cofactor expansion of the determinant along the first row, since det A is written
as the sum of the products of each entry of the first row and the corresponding
cofactor (i.e., the sum of the terms a\j times (— |Ay |).
It is natural to ask if one can compute determinants by cofactor expansion
along other rows or columns of A. Happily, the answer is yes (although we shall
not prove this).
58 Chapter 1 | Vectors
Convenient Fact. The determinant of A can be computed by cofactor ex-
pansion along any row or column. That is,
\A\ = (-ly+'anl An | + (-l)i+2ai2\Ai2\ + ■ ■
■ + (-i)i+nain\Ain\
(expansion along the rth row),
|A| = (-l)!+^ly.|Aly-| + (-l)2+Ja2j \A2j\ + •
■ + {-X)n+>anl\Anj\
(expansion along the y'th column).
EXAMPLE 9 To compute the determinant of
- 1
2
0
4
5 _
2
0
0
9
0
7
5
1
-1
0
0
2
0
0
2
3
1
0
0
0
expansion along the first row involves more calculation than necessary. In partic-
ular, one would need to calculate four 4x4 determinants on the way to finding
the desired 5x5 determinant. (To make matters worse, these 4x4 determinants
would in turn, need to be expanded also.) However, if we expand along the third
column, we find that
det A = (- l)1+3(0) det A13 + (- 1)2+3(0) det A23 + (- 1)3+3(1) det A33
+ (- 1)4+3(0) det A43 + (- 1)5+3(0) det A53
= det A33
1
2
4
5
2
0
9
0
0
2
0
2
3
1
0
0
There are several good ways to evaluate this 4x4 determinant. We'll expand
about the bottom row:
12 4 5
2 0 9 0
0 2 0 2
3 10 0
2
4 5
1
4 5
(-1)4+1(3)
0
9 0
+ (-l)4+2(l)
2
9 0
2
0 2
0
0 2
= (-l)(3)(-54) + (l)(l)(2)
= 164.
Of course, not all matrices contain well-distributed zeros as in Example 9,
so there is by no means always an obvious choice for an expansion that avoids
much calculation. Indeed one does not compute determinants of large matrices
by means of cofactor expansion. Instead certain properties of determinants are
used to make hand computations feasible. Since we shall rarely need to consider
determinants larger than 3 x 3, we leave such properties and their significance to
the exercises. (See, in particular, Exercises 26 and 27.)
1.6 | Exercises 59
1.6 Exercises
1 . Rewrite in terms of the standard basis for R" :
(a) (1,2,3, ...,n)
(b) (1,0,-1, 1,0,-1,..., 1,0, -1) (Assume that n
is a multiple of 3.)
In Exercises 2—4 write the given vectors without recourse to
standard basis notation.
2. ei + e2 H he,,
3. ei - 2e2 + 3e3 - 4e4 H h (-l)"+1ne„
4. ei + e„
5. Calculate the following, where a = (1, 3, 5, . . . ,
In - 1) and b = (2, -4, 6, ... , (-l)"+12w):
(a) a + b
(d) ||a||
(b) a - b
(e) a-b
(c) -3a
6. Let n be an even number. Verify the triangle in-
equality in R" for a = (1, 0, 1, 0, ... , 0) and b =
(0, 1, 0, 1, ... , 1).
7. Verify that the Cauchy-Schwarz inequality holds for
the vectors a = (1 , 2, . . . , n) and b = ( 1 , 1 , . . . , 1).
8. If a = (1,-1,7, 3, 2) and b = (2, 5, 0, 9, -1), calcu-
late the projection projab.
9. Show, for all vectors a, b, c e R", that
|| a — b|| < || a — c|| + ||c — b||.
10. Prove the Pythagorean theorem. That is, if a, b, and
c are vectors in R" such that a + b = c and a • b = 0,
then
||a||2 + ||b||2 = ||c||2.
Why is this called the Pythagorean theorem?
11. Let a and b be vectors in R". Show that if ||a + b|| =
|| a — b||, then a and b are orthogonal.
12. Let a and b be vectors in R". Show that if ||a — b|| >
|| a + b || , then the angle between a and b is obtuse (i.e.,
more than jt/2).
13. Describe "geometrically" the set of points in R5 satis-
fying the equation
2(Xl - 1) + 3(x2 + 2) - 7x3 + x4 - 4 - 5(x5 + 1) = 0.
14. To make some extra money, you decide to print four
types of silk-screened T-shirts that you sell at various
prices. You have an inventory of 20 shirts that you can
sell for $8 each, 30 shirts that you sell for $10 each, 24
shirts that you sell for $ 12 each, and 20 shirts that you
sell for $ 1 5 each. A friend of yours runs a side business
selling embroidered baseball caps and has an inventory
of 30 caps that can be sold for $10 each, 16 caps that
can be sold for $10 each, 20 caps that can be sold for
$12 each, and 28 caps that can be sold for $15 each.
You suggest swapping half your inventory of each type
of T-shirt for half his inventory of each type of baseball
cap. Is your friend likely to accept your offer? Why or
why not?
15. Suppose that you run a grain farm that produces six
types of grain at prices of $200, $250, $300, $375,
$450, $500 per ton.
(a) If x = (xi, . . . , xg) is the commodity bundle vec-
tor (meaning that x,- is the number of tons of grain
i to be purchased), express the total cost of the
commodity bundle as a dot product of two vectors
inR6.
(b) A customer has a budget of $100,000 to be used
to purchase your grain. Express the set of possible
commodity bundle vectors that the customer can
afford. Also describe the relevant budget hyper-
plane in R6.
In Exercises 16-19, calculate the indicated matrix quantities
where
C
i :
-2 (
1 -1
2 0
0 3
16. 3A-2S
18. DB
, D =
17. AC
19. BTD
20. The n x n identity matrix, denoted I or 7„, is the ma-
trix whose nth entry is 1 and whose other entries are
all zero. That is,
0 0
(a) Explicitly write out I2, h, and I4.
(b) The reason / is called the identity matrix is that
it behaves as follows: Let A be any m x n matrix.
Then
i. AI„ = A.
ii. ImA = A.
Prove these results. (Hint: What are the yth entries
of the products in (i) and (ii)?)
Chapter 1 | Vectors
Evaluate the determinants given in Exercises 21—23.
22.
-7
/
A
u
1
— 1
0
2
0
1
3
1
-3
0
2
0
5
1
-2
o
0
A
A
U
0
1 c
1
1
A
u
0
7
0
1
— 1
0
8
1
9
7
5
-1
0
8
11
0
2
1
9
7
0
0
4
-3
5
0
0
0
2
1
0
0
0
0
-3
24. Prove that a matrix that has a row or a column con-
sisting entirely of zeros has determinant equal to
zero.
25. An upper triangular matrix is an n x n matrix whose
entries below the main diagonal are all zero. (Note:
The main diagonal is the diagonal going from upper
left to lower right.) For example, the matrix
- 1
2
-1
2 "
0
3
4
3
0
0
5
6
_ 0
0
0
7 _
is upper triangular.
(a) Give an analogous definition for a lower triangu-
lar matrix and also an example of one.
(b) Use cofactor expansion to show that the determi-
nant of any n x n upper or lower triangular matrix
A is the product of the entries on the main diagonal.
That is, det A = awa^i ■ ■ ■ a„„ .
26. Some properties of the determinant. Exercises 24
and 25 show that it is not difficult to compute de-
terminants of even large matrices, provided that the
matrices have a nice form. The following operations
(called elementary row operations) can be used to
transform annxn matrix into one in upper triangular
form:
I. Exchange rows i and j.
II. Multiply row by a nonzero scalar.
III. Add a multiple of row i to row j. (Row i remains
unchanged.)
For example, one can transform the matrix
" 0 2 3 "
1 7 -2
_ 1 5 9 _
into one in upper triangular form in three steps:
Step 1 . Exchange rows 1 and 2 (this puts a nonzero
entry in the upper left corner):
0
2
3 "
1
7
-2
1
7
-2
0
2
3
1
5
9
1
5
9
Step 2. Add —1 times row 1 to row 3 (this eliminates
the nonzero entries below the entry in the upper left
corner):
Step 3
1
7 -2
" 1
7
-2
0
2 3
0
2
3
1
5 9
0
-2
11
Add row 2 to row 3:
1
7 -2
" 1
7
-2
0
2 3
0
2
3
0
-2 11
0
0
14
The question is, how do these operations affect the de-
terminant?
(a) By means of examples, make a conjecture as to the
effect of a row operation of type I on the determi-
nant. (That is, if matrix B results from matrix A by
performing a single row operation of type I, how are
det A and det B related?) You need not prove your
results are correct.
(b) Repeat part (a) in the case of a row operation of
type III.
(c) Prove that if B results from A by multiplying the
entries in the ;'th row of A by the scalar c (a type II
operation), then det B = c ■ det A.
27. Calculate the determinant of the matrix
" 2
1
-2
7
8
1
0
1
-2
4
-1
1
2
3
-5
0
2
3
1
7
_ -3
2
-1
0
1
by using row operations to transform A into a matrix
in upper triangular form and by using the results of
Exercise 26 to keep track of how the determinant of A
and the determinant of your final matrix are related.
28. (a) Is det(A + B) = det A + det 5? Why or why not?
(b) Calculate
1 2 7
3+2 1-1 5+1
0-2 0
and
1
2
7
1
2
7
3
1
5
+
2
-1
1
0
-2
0
0
-2
0
and compare your results.
1.6 | Exercises 61
1
3
2 + 3
0
4
-1+5
-1
0
0-2
1
3
2
1
3
3
0
4
-1
+
0
4
5
-1
0
0
-1
0
-2
(c) Calculate
and
and compare your results.
(d) Conjecture and prove a result about sums of deter-
minants. (You may wish to construct further exam-
ples such as those in parts (b) and (c).)
29. It is a fact that, if A and B are any n x n matrices, then
det(AB) = (detA)(detfi).
Use this fact to show that det(A5) = det(BA). (Recall
that AB / BA, in general.)
An n x n matrix A is said to be invertible (or nonsingular) if
there is another n x n matrix B with the property that
AB = BA = I„,
where In denotes the n x n identity matrix. (See Exercise 20.)
The matrix B is called an inverse to the matrix A. Exercises
30-38 concern various aspects of matrices and their inverses.
30. (a) Verify that
(b) Verify that
1 0
1 1
is an inverse of
1 0
-1 1
1 2 3
2 5 3
1 0 8
is an inverse of
-40 16 9
13 -5 -3
5 -2 -1
31. Using the definition of an inverse matrix, find an
2 2 1
inverse to 0 1 0
0 0-1
32. Try to find an inverse matrix to
What happens?
2 1
1 0
0 -1
33. Show that if an n x n matrix A is invertible, then A can
have only one inverse matrix. Thus, we may write A-1
to denote the unique inverse of a nonsingular matrix
A. (Hint: Suppose A were to have two inverses B and
C. Consider B(AC).)
34. Suppose that A and B are n x « invertible matrices.
Show that the product matrix A 5 is invertible by ver-
ifying that its inverse (AS)-1 = B~l A~l .
35. (a) Show that if A is invertible, then det A / 0. (In
fact, the converse is also true.)
(b) Show that if A is invertible, then
det(A"1)= — — .
v ; detA
36. (a) Show that, if ad — be / 0, then a general 2x2
a b
matrix
1
c d
has the matrix
ad — be
d
-c
ad— be
c
ad— be
as inverse.
(b) Use this formula to find an inverse of
ad— be
a
ad— be
2 4
-1 2
37. If A is a 3 x 3 matrix and det A / 0, then there is
a (somewhat complicated) formula for A-1. In
particular,
detA
|A„|
-IA12I
|An|
-IA21I
IA22I
-IA23I
IA31I
-IA32I
IA33I
where Ay denotes the submatrix of A obtained by
deleting the ;th row and 7th column (see Defini-
tion 6.5). Use this formula to find the inverse of
2 1 1
0 2 4
1 0 3
More generally, if A is any n x n matrix and det A^0,
then
detA
where adj A is the adjoint matrix of A, that is, the
matrix whose ijth entry is (—\)1+-'\Aji\. (Note: The
formula for the inverse matrix using the adjoint is typi-
cally more of theoretical than practical interest, as there
are more efficient computational methods to determine
the inverse, when it exists.)
38. Repeat Exercise 37 with the matrix
2-1 3
1 2 -2
3 0 1
Cross products in R". Although it is not possible to define a
cross product of two vectors in R" as we did for two vectors
in R3, we can construct a "cross product" of n — 1 vectors
in R" that behaves analogously to the three-dimensional cross
62 Chapter 1 [ Vectors
product. To be specific, if
ai = (an, au, ain), a2 = (021, 022, ■ ■ • , a2„),
= (fln-u, in-12, ■ ■ ■ , a„-\„)
are n — 1 vectors in R", we define ai X a2 X • • • X a„_i to be
the vector in R" given by the symbolic determinant
ai x a2 x • • • x a„_i
"21
a„-i 1
e2
"12
"22
a„-i2
e„
«2»
(Here t\ , . . . , e„ are £/ze standard basis vectors for R" .) Exer-
cises 39-42 concern this generalized notion of cross product.
39. Calculate the following cross product in R4:
(1, 2, -1, 3) x (0, 2, -3, 1) x (-5, 1, 6, 0).
40. Use the results of Exercises 26 and 28 to show that
(a) ai x • • • x a,- x • • • x a^ x • • • x a„_i
= — (ai x • • • x a j x • • • x a,- x • • • x a„_i ),
I <i <n — 1 , 1 < j < n — 1
(b) ai x • • • x kHi x • • • x a„_i
= &(ai x • • • x a,- x • • • x a„_i),
1 < i < n — 1 .
(c) ai x • • • x (a,- + b) x • • • x a„_i
= ai x • • • x a,- x • • • x a„_i +
aix---xbx---x a„_i,
1 <i <n - l.allbeR".
(d) Show that if b = (b\ , . . . , b„) is any vector in R" ,
then
b • (ai x a2 x • • • x a„_i)
is given by the determinant
All
aB-n
b„
41. Show that the vector b = ai x a2 x • • • x a„_i is
orthogonal to ai , . . . , a„_i .
42. Use the generalized notion of cross products to
find an equation of the (four-dimensional) hyper-
plane in R5 through the five points Po(l, 0, 3, 0, 4),
^(2,-1,0,0,5), P2(7, 0, 0, 2, 0), P3(2,0,3,0,4),
and P4(l,~ 1,3,0, 4).
Figure 1 .82 The Cartesian
coordinate system.
Location y
(x,y)
Location x
Figure 1 .83 Locating a point P,
using Cartesian coordinates.
1 .7 New Coordinate Systems
We hope that you are comfortable with Cartesian (rectangular) coordinates for R2
or R3 . The Cartesian coordinate system will continue to be of prime importance to
us, but from time to time, we will find it advantageous to use different coordinate
systems. In R2, polar coordinates are useful for describing figures with circular
symmetry. In R3, there are two particularly valuable coordinate systems besides
Cartesian coordinates: cylindrical and spherical coordinates. As we shall see,
cylindrical and spherical coordinates are each a way of adapting polar coordinates
in the plane for use in three dimensions.
Cartesian and Polar Coordinates on R2
You can understand the Cartesian (or rectangular) coordinates (x , y) of a point
P in R2 in the following way: Imagine the entire plane filled with horizontal and
vertical lines, as in Figure 1.82. Then the point P lies on exactly one vertical line
and one horizontal line. The x -coordinate of P is where this vertical line intersects
the x-axis, and the y-coordinate is where the horizontal line intersects the y-axis.
(See Figure 1.83.) (Of course, we've already assigned coordinates along the axes
so that the zero point of each axis is at the point of intersection of the axes. We
also normally mark off the same unit distance on each axis.) Note that, because
of this geometry, every point in R2 has a uniquely determined set of Cartesian
coordinates.
Polar coordinates are defined by considering different geometric informa-
tion. Now imagine the plane filled with concentric circles centered at the origin
and rays emanating from the origin. Then every point except the origin lies on
1.7 | New Coordinate Systems 63
Figure 1 .86 Locating the point
with polar coordinates (r, 9),
where r < 0.
0
r = 6 cos 0
0
6
jr/6
3V3
n/4
3V2
7T/3
3
n/2
0
2jt/3
-3
3jt/4
-3V2
5?r/6
-3V3
7T
-6
77T/6
-3V3
57T/4
-3V2
4tt/3
-3
3^/2
0
57T/3
3
7tt/4
372
Figure 1 .84 The polar coordinate
system.
Figure 1 .85 Locating a point P,
using polar coordinates.
exactly one such circle and one such ray. The origin itself is special: No circle
passes through it, and all the rays begin at it. (See Figure 1 .84.) For points P other
than the origin, we assign to P the polar coordinates (r, 9), where r is the radius
of the circle on which P lies and 9 is the angle between the positive x-axis and
the ray on which P lies. (9 is measured as opening counterclockwise.) The origin
is an exception: It is assigned the polar coordinates (0, 9), where 9 can be any
angle. (See Figure 1.85.) As we have described polar coordinates, r > 0 since r
is the radius of a circle. It also makes good sense to require 0 < 9 < 2jt, for then
every point in the plane, except the origin, has a uniquely determined pair of polar
coordinates. Occasionally, however, it is useful not to restrict r to be nonnegative
and 9 to be between 0 and lit. In such a case, no point of R2 will be described by
a unique pair of polar coordinates: If P has polar coordinates (r, 9), then it also
has (r, 9 + 2nn) and (— r, 9 + (2n + \)tt) as coordinates, where n can be any
integer. (To locate the point having coordinates (r, 9), where r < 0, construct the
ray making angle 9 with respect to the positive x-axis, and instead of marching
| r | units away from the origin along this ray, go | r \ units in the opposite direction,
as shown in Figure 1.86.)
EXAMPLE 1 Polar coordinates may already be familiar to you. Nonetheless,
make sure you understand that the points pictured in Figure 1.87 have the coor-
dinates indicated. ♦
(3V3, kI6)
(2, 5n/6)
(2, kI6)
(5,0)
(-1, 5tt/6) or
(1, 1171/6) or
(3, 3tt/2) <
(l,-7i/6)
Figure 1 .87 Figure for
Example 1.
Figure 1 .88 The graph of
r = 6 cos 6 in Example 2.
EXAMPLE 2 Let's graph the curve given by the polar equation r = 6cos#
(Figure 1.88). We can begin to get a feeling for the graph by compiling values, as
in the adjacent tabulation.
64 Chapter 1 [ Vectors
Thus, r decreases from 6 to 0 as 9 increases from 0 to ji/2; r decreases from
0 to —6 (or is not defined if you take r to be nonnegative) as 9 varies from ir/2
to 7r; r increases from —6 to 0 as 9 varies from n to 3jt/2; and r increases from
0 to 6 as 9 varies from 3jt/2 to 2ir . To graph the resulting curve, imagine a radar
screen: As 9 moves counterclockwise from 0 to 2n, the point (r, 9) of the graph is
traced as the appropriate "blip" on the radar screen. Note that the curve is actually
traced twice: once as 9 varies from 0 to n and then again as 9 varies from it to
2jt. Alternatively, the curve is traced just once if we allow only 9 values that yield
nonnegative r values. The resulting graph appears to be a circle of radius 3 (not
centered at the origin), and in fact, one can see (as in Example 3) that the graph
is indeed such a circle. ♦
The basic conversions between polar and Cartesian coordinates are provided
by the following relations:
Polar to Cartesian:
J x = r cos9 ,
1 y = r sin 9 '
Cartesian to polar:
\ r2 = x2 + y2
(tan<9 = y/x - (2)
Note that the equations in (2) do not uniquely determine r and 9 in terms of x
and y. This is quite acceptable, really, since we do not always want to insist that
r be nonnegative and 9 be between 0 and 2n. If we do restrict r and 9, however,
then they are given in terms of x and y by the following formulas:
r = y/x2 + y2,
tan-1 y/x
if x
>
0,y
>
0
tan-1 y/x + 2jt
ifx
>
0,y
<
0
tan-1 y/x + n
ifx
<
0,y
>
0
tt/2
ifx
0,y
>
0
3tt/2
ifx
0,y
<
0
indeterminate
ifx
y =
0
The complicated formula for 9 arises because we require 0 < 9 < 2n , while the
inverse tangent function returns values between — ir/2 and n/2 only. Now you
see why the equations given in (2) are a better bet!
EXAMPLE 3 We can use the formulas in (1) and (2) to prove that the curve in
Example 2 really is a circle. The polar equation r = 6 cos 9 that defines the curve
requires a little ingenuity to convert to the corresponding Cartesian equation. The
trick is to multiply both sides of the equation by r. Doing so, we obtain
r2 = 6r cosO.
Now (1) and (2) immediately give
x2 + y2 = 6x.
1.7 | New Coordinate Systems 65
We complete the square in x to find that this equation can be rewritten as
(x - 3)2 + y2 = 9,
which is indeed a circle of radius 3 with center at (3, 0). ♦
Cylindrical Coordinates
Cylindrical coordinates on R3 are a "naive" way of generalizing polar coordinates
to three dimensions, in the sense that they are nothing more than polar coordinates
used in place of the x- and y-coordinates. (The z-coordinate is left unchanged.)
The geometry is as follows: Except for the z-axis, fill all of space with infinitely
extended circular cylinders with axes along the z-axis as in Figure 1.89. Then
any point P in R3 not lying on the z-axis lies on exactly one such cylinder.
Hence, to locate such a point, it's enough to give the radius of the cylinder, the
circumferential angle 9 around the cylinder, and the vertical position z along the
cylinder. The cylindrical coordinates of P are (r, 9, z), as shown in Figure 1.90.
Algebraically, the equations in (1) and (2) can be extended to produce the basic
conversions between Cartesian and cylindrical coordinates.
The basic conversions between cylindrical
and Cartesian coordinates
are
provided by the following relations:
x = r cos 6*
Cylindrical to Cartesian:
y = r sin 9 ;
Z = Z
(3)
r2 = x2 + y2
Cartesian to cylindrical:
tan 9 = y/x .
(4)
z = z
As with polar coordinates, if we make the restrictions r > 0, 0 < 9 < 2n, then
all points of R3 except the z-axis have a unique set of cylindrical coordinates. A
point on the z-axis with Cartesian coordinates (0,0, zo) has cylindrical coordinates
(0, 9, zo), where 9 can be any angle.
Cylindrical coordinates are useful for studying objects possessing an axis of
symmetry. Before exploring a few examples, let's understand the three "constant
coordinate" surfaces.
• The r = T(, surface is, of course, just a cylinder of radius ro with axis the
z-axis. (See Figure 1.91.)
• The 9 = 9q surface is a vertical plane containing the z-axis (or a half-plane
with edge the z-axis if we take r > 0 only). (See Figure 1.92.)
• The z = zo surface is a horizontal plane. (See Figure 1.93.)
Half-plane only
EXAMPLE 4 Graph the surface having cylindrical equation r = 6 cos 8. (This
equation is identical to the one in Example 2.) In particular, z does not appear
in this equation. What this means is that if the surface is sliced by the horizontal
plane z = c where c is a constant, we will see the circle shown in Example 2, no
matter what c is. If we stack these circular sections, then the entire surface is a
circular cylinder of radius 3 with axis parallel to the z-axis (and through the point
(3, 0, 0) in cylindrical coordinates). This surface is shown in Figure 1.94. ♦
EXAMPLE 5 Graph the surface having equation z = 2r in cylindrical co-
ordinates.
Here the variable 6 does not appear in the equation, which means that the
surface in question will be circularly symmetric about the z-axis. In other words,
if we slice the surface by any plane of the form 6 = constant (or half-plane, if we
take r > 0), we see the same curve, namely, a line (respectively, a half-line) of
slope 2. As we let the constant-0 plane vary, this line generates a cone, as shown
in Figure 1 .95. The cone consists only of the top half (nappe) when we restrict r
to be nonnegative.
The Cartesian equation of this cone is readily determined. Using the formulas
in (4), we have
Z = 2r => z2=4r2 <^=> z2 = 4(x2 + y2).
Since z can be positive as well as negative, this last Cartesian equation describes
the cone with both nappes. If we want the top nappe only, then the equation
z = 2y 'x2 + y2 describes it. Similarly, z = —2^x2 + y2 describes the bottom
nappe. ♦
Spherical Coordinates
Fill all of space with spheres centered at the origin as in Figure 1 .96. Then every
point P e R3, except the origin, lies on a single such sphere. Roughly speaking,
the spherical coordinates of P are given by specifying the radius p of the sphere
containing P and the "latitude and longitude" readings of P along this sphere.
More precisely, the spherical coordinates (p, <p, 9) of P are defined as follows: p
is the distance from P to the origin; <p is the angle between the positive z-axis and
the ray through the origin and P; and 6 is the angle between the positive x-axis
and the ray made by dropping a perpendicular from P to the xy-plane. (See Figure
1 .97.) The 0 -coordinate is exactly the same as the ^-coordinate used in cylindrical
coordinates. (Warning: Physicists usually prefer to reverse the roles of <p and 9,
as do some graphical software packages.)
1.7 | New Coordinate Systems 67
z
Figure 1 .96 The spherical Figure 1 .97 Locating the point
coordinate system. P, using spherical coordinates.
It is standard practice to impose the following restrictions on the range of
values for the individual coordinates:
P > 0, 0 < <p < n, 0 < 9 < lit. (5)
With such restrictions, all points of R3 , except those on the z-axis, have a uniquely
determined set of spherical coordinates. Points along the z-axis, except for the
origin, have coordinates of the form (po, 0, 9) or (po, n, 9), where po is a positive
constant and 9 is arbitrary. The origin has spherical coordinates (0,<p,9), where
both (p and 6 are arbitrary.
EXAMPLE 6 Several points and their corresponding spherical coordinates are
shown in Figure 1.98. ♦
Figure 1.98 Figure for Example 6. Figure 1.99 The graph of Figure 1.100 The spherical
P = Po (> 0). surface <p = (po, shown for
different values of (po.
Spherical coordinates are especially useful for describing objects that have
a center of symmetry. With the restrictions given by the inequalities in (5), the
constant coordinate surface p = po (po > 0) is, of course, a sphere of radius po,
as shown in Figure 1.99. The surface given by 9 = 9o is a half-plane just as in the
cylindrical case. The <p = <po surface is a single-nappe cone if cpo / tt/2 and is
the xy-plane if cp0 = tt/2 (and is the positive or negative z-axis if (po = 0 or it).
(See Figure 1 . 1 00.) If we do not insist that p be nonnegative, then the cones would
include both nappes.
68 Chapter 1 [ Vectors
The basic equations relating spherical coordinates to both cylindrical and
Cartesian coordinates are as follows.
Spherical/cylindrical:
r = p sin cp
—2 2 i _2
p = r + z
9 = 9
\my = r/z . (6)
z = p cos cp
9=9
Spherical/Cartesian:
x = psincp cos 9
p2 = x2 + y2 + z2
y = p sin (p sin 9
tan<p = y/x2 + y2/z- (7)
z = p coscp
tan# = y/x
Figure 1.101 Converting
spherical to cylindrical coordinates
when 0 < <p < j.
Figure 1.102 Converting
spherical to cylindrical
coordinates when jt/2 < cp < it .
Using basic trigonometry, it is not difficult to establish the conversions in (6).
From the right triangle shown in Figure 1.101, we have
COS I
Hence,
Similarly,
so that
r = p cos
p sin<p.
sin
p sin
(*-')-
(£-')-
z
P
p cos (p.
Thus, the formulas in (6) follow when 0 < <p < tt/2. If it/2 < <p < n, then we
may employ Figure 1.102. So
p cos
('-I)
p sm<p,
and
z = -P sin ^ - -) = p sin - <p) =
p cos<p.
1.7 | New Coordinate Systems 69
V
p = 2a cos <p
0
2a
7T/6
V3a
ji/4
V2a
tt/3
a
JT/2
0
2jt/3
—a
3jt/4
— \/2a
7T
—2a
Hence, the relations in (6) hold in general. The equations in (7) follow by substi-
tution of those in (6) into those of (3) and (4).
EXAMPLE 7 The cylindrical equation z = 2r in Example 5 converts via (6)
to the spherical equation
p cosco = 2p simp.
Therefore,
tano5
26°.
1 -i 1
- <=?■ co = tan -
2 r 2
Thus, the equation defines a cone (as we just saw). The spherical equation is
especially simple in that it involves just a single coordinate. ♦
EXAMPLE 8 Not all spherical equations are improvements over their cylindri
y
cal or Cartesian counterparts. For example, the Cartesian equation 6x = x2 1
(whose polar-cylindrical equivalent is r = 6 cos 9) becomes
6p sin (p cos 9 = p2 sin2 05 cos2 9 + p2 sin2 cp sin2 9
from (7). Simplifying,
6p sin (p cos 9 = p sin <p (cos © + sin 9)
6p sin 09 cos 9 = p2 sin2 <p
6cos# = p sin 05.
This spherical equation is more complicated than the original Cartesian equation
in that all three spherical coordinates are involved. Therefore, it is not at all
obvious that the spherical equation describes a cylinder. ♦
EXAMPLE 9 Let's graph the surface with spherical equation p = 2a cosoo,
where a > 0. As with the graph of the cone with cylindrical equation z = 2r,
note that the equation is independent of 9. Thus, all sections of this surface made
by slicing with the half-plane 9 = c must be the same. If we compile values as
in the adjacent table, then the section of the surface in the half-plane 9 = 0 is as
shown in Figure 1.103. Since this section must be identical in all other constant-0
half-planes, we see that this surface appears to be a sphere of radius a tangent to
the xy-plane, which is shown in Figure 1.104.
(2a, 0, 0)
Section of
p = 2a cos q>
Figure 1.103 The cross section
of p = 2a cos 05 in the half-plane
6 = 0.
Figure 1 .1 04 The graph of p =
2a cos<p.
Chapter 1 | Vectors
The Cartesian equation of the surface is determined by multiplying both sides
of the spherical equation by p and using the conversion equations in (7):
p = 2a cos <p p2 = lap cos <p
<^=> x2 + y2 + z2 = 2az
<^=> x2 + y2 + (z - a)2 = a2
by completing the square in z. This last equation can be recognized as that of a
sphere of radius a with center at (0, 0, a) in Cartesian coordinates. ♦
EXAMPLE 10 NASA launches a 10-ft-diameter space probe. Unfortunately,
a meteor storm pushes the probe off course, and it is partially embedded in
the surface of Venus, to a depth of one quarter of its diameter. To attempt to
reprogram the probe's on-board computer to remove it from Venus, it is necessary
to describe the embedded portion of the probe in spherical coordinates. Let us
find the description desired, assuming that the surface of Venus is essentially flat
in relation to the probe and that the origin of our coordinate system is at the center
of the probe.
z = -5/2
Figure 1 .1 07 Coordinate view of
the cross section of the probe of
Example 10.
The situation is illustrated in Figure 1 . 1 05 . The buried part of the probe clearly
has symmetry about the z-axis. That is, any slice by the half-plane 9 = constant
looks the same as any other. Thus, 9 can vary between 0 and 2it . A typical slice
of the probe is shown in Figure 1 . 106. Elementary trigonometry indicates that for
the angle a in Figure 1.106,
1 1
cos a = — = -.
5 2
Hence, a = cos-1 \ = n/3. Thus, the spherical angle <p (which opens from the
positive z-axis) varies from n — tt/3 = 2n/3 to it as it generates the buried part
of the probe. Finally, note that for a given value of y> between 2n/3 and it, p
is bounded by the surface of Venus (the plane z = — | in Cartesian coordinates)
and the spherical surface of the probe (whose equation in spherical coordinates is
p = 5). See Figure 1.107. From the formulas in (7) the equation z = — f corre-
sponds to the spherical equation p cos (p = — | or, equivalently, to p = — | sec <p.
Therefore, the embedded part of the probe may be defined by the set
5 2tt „ '
- seccp < p < 5, — <(p<tt,0<9<2tt
1.7 | New Coordinate Systems
Standard Bases for Cylindrical
and Spherical Coordinates
In Cartesian coordinates, there are three special unit vectors i, j, and k that point
in the directions of increasing x-, y-, and z-coordinate, respectively. We find
corresponding sets of vectors for cylindrical and spherical coordinates. That is,
in each set of coordinates, we seek mutually orthogonal unit vectors that point in
the directions of increasing coordinate values.
In cylindrical coordinates, the situation is as shown in Figure 1.108. The
vectors e, , ee, and e-, which form the standard basis for cylindrical coordinates,
are unit vectors that each point in the direction in which only the coordinate
indicated by the subscript increases. There is an important difference between
the standard basis vectors in Cartesian and cylindrical coordinates. In the former
case, i, j, and k do not vary from point to point. However, the vectors e, and eg
do change as we move from point to point.
Now we give expressions for er, eg, and ez. Since the cylindrical z-coordinate
is the same as the Cartesian z-coordinate, we must have ez = k. The vector
er must point radially outward from the z-axis with no k-component. At a
point (x, y, z) 6 R3 (Cartesian coordinates), the vector x'\ + y\ has this prop-
erty. Normalizing it to obtain a unit vector (see Proposition 3.4 of §1.3), we
obtain
= xi + y\
v7*2 + y2 '
With er and e- in hand it's now a simple matter to define eg, since it must be
perpendicular to both e,- and ez. We take
-yi + xj
eg = e;xer =
V*2 + y2
(The reason for this choice of cross product, as opposed to e, x ez, is so that eg
points in the direction of increasing 0.) To summarize, and using the cylindrical
to Cartesian conversions given in (3),
e,
xi + yj
= — = cos# i + sm6 j;
V*2 + y2
eg
-yi + *j . a. . a.
= — = — sin0 1 + cos0 j;
V*2 + v2
(8)
= k.
In spherical coordinates, the situation is shown in Figure 1 . 109. In particular,
there are three unit vectors ep , , and eg that form the standard basis for spherical
coordinates. These vectors all change direction as we move from point to point.
We give expressions for ep , , and ee . Since the ^-coordinates in both spher-
ical and cylindrical coordinates mean the same thing, eg in spherical coordinates
is given by the value of eg in (8). At a point (x,y, z), the vector ep should point
72 Chapter 1 [ Vectors
from the origin directly to (x, y, z). Thus, ep may be obtained by normalizing
■*i + yi + zk. Finally, ep is nothing more than eg x e^. If we explicitly perform
the calculations just described and make use of the conversion formulas in (7),
the following are obtained:
xi + yj + zk
= — = sm<p cos 6
Vx2 + y2 + z2
i + sin <p sin 0 j + cos <p k;
xzi + yzj - (x2 + y2)k
v7*2 + y2 Vx2 + y2 + ^2
(9)
sin <p k;
= cos <p cos 0 i + cos cp sin 0 j -
— yi + xj
= — = — sm 6 1 + cos 6
V*2 +
Although the results of (8) and (9) will not be used frequently, they will prove
helpful on occasion.
Hyperspherical Coordinates (optional)
There is a way to provide a set of coordinates for R" that generalizes spherical
coordinates on R3. For n > 3, the hyperspherical coordinates of a point P e R"
are (p, q>\, q>%, . . . , <pn-i) and are defined by their relations with the Cartesian
coordinates (xi, x%, . . . , xn) of P as
' xi = p sin sin ip2 ■ ■ ■ sin <p„_2 cos <p„_ i
x2 = p sin^i sin ip2 ■ ■ ■ sin<p„_2 sin<pn_i
x3 = p sinip! sin • • • sin<p,,_3 cos<p„_2
(10)
X4 = p sm ipi sm q>2 - - ■ sm <p„_4 cos <p„_3
_xn = p COS^i
To be more explicit, in equation (10) above we take
X* = p sin sin cp2 - ■ ■ sin cos ^„_jt+i for k = 3, . . . , n.
Note that when n = 3, the relations in (10) become
xi = p sin^i cos i^2
• X2 = p sin^i sini£>2 ■
*3 = pcos^i
These relations are the same as those given in (7), so hyperspherical coordinates
are indeed the same as spherical coordinates when n = 3.
In analogy with (5), it is standard practice to impose the following restrictions
on the range of values for the coordinates:
P > 0, 0 < (pi < 7T for k = 1, . . . , n — 2, 0 < q>n-\ < lit. (1 1)
1.7 | Exercises 73
Then, with these restrictions, we can convert from hypersphencal coordinates to
Cartesian coordinates by means of the following formulas:
' p2 = x\ + x\ H hi„2
tanpi = ^2 H hx2_,/x„
tan ^2 = J A H h x*_2/Xn-i n 9,
tan<p„_2 = J + *f/*3
tan^„_! = x2/*i
Hyperspherical coordinates get their name from the fact that the (n — 1)-
dimensional hypersurface in R" defined by the equation p = p0, where po is a
positive constant, consists of points on the hypersphere of radius po defined in
Cartesian coordinates by the equation
x2+x2 + ---+x2 = p2.
1.7 Exercises
In Exercises 1—3, find the Cartesian coordinates of the points
whose polar coordinates are given.
1. (V2,jr/4)
2. (V3, 5jt/6)
3. (3, 0)
In Exercises 4-6, give a set of polar coordinates for the point
whose Cartesian coordinates are given.
4. (273,2)
5. (-2,2)
6. (-1,-2)
In Exercises 7-9, find the Cartesian coordinates of the points
whose cylindrical coordinates are given.
7. (2,2,2)
8. (n,n/2, 1)
9. (1,2* /3, -2)
In Exercises 10-13, find the rectangular coordinates of the
points whose spherical coordinates are given.
10. (4, ;r/2,7r/3)
11. (3, ;r/3,7r/2)
12. (1,3tt/4, 2jt/3)
13. (2,;r,7r/4)
In Exercises 14-16, find a set of cylindrical coordinates of the
point whose Cartesian coordinates are given.
14. (-1,0,2)
15. (-1, V3, 13)
16. (5,6,3)
In Exercises 1 7 and 18, find a set of spherical coordinates of
the point whose Cartesian coordinates are given.
17. (1, -1, v/6)
18. (0, V3, 1)
19. This problem concerns the surface described by the
equation (r — 2)2 + z2 = 1 in cylindrical coordinates.
(Assume r > 0.)
(a) Sketch the intersection of this surface with the half-
plane 8 = jt/2.
(b) Sketch the entire surface.
20. (a) Graph the curve in R2 having polar equation r =
2a sin 9, where a is a positive constant.
(b) Graph the surface in R3 having spherical equation
p = 2a sin <p.
21. Graph the surface whose spherical equation is p =
1 — cosip.
22. Graph the surface whose spherical equation is p =
1 — sinip.
In Exercises 23-25, translate the following equations from
the given coordinate system (i.e., Cartesian, cylindrical, or
74 Chapter 1 | Vectors
spherical) into equations in each of the other two systems. In
addition, identify the surfaces so described by providing ap-
propriate sketches.
23. p sin<p sin# = 2
24. z2 = 2x2 + 2y2
25. r = 0
In Exercises 26-29, sketch the solid whose cylindrical coordi-
nates (r, 9, z) satisfy the given inequalities.
26. 0<r<3, 0<6»<tt/2, -1 < z < 2
27. r < z < 5, 0 < 9 < n
28. 2r < z < 5 - 3r
29. r2 - 1 < z < 5 - r2
In Exercises 30-35, sketch the solid whose spherical coordi-
nates (p, <p, 9) satisfy the given inequalities.
30. 1 < p < 2
31. 0 < p < 1, 0 < (p < tt/2
32. 0 < p < 1, 0 < 6» < jr/2
33. 0 < p < jt/4, 0 < p < 2
34. 0 < p < 2/coscp, 0 < tp < tt/4
35. 2 cos (p < p <3
36. (a) Which points P in R2 have the same rectangular
and polar coordinates?
(b) Which points P in R3 have the same rectangular
and cylindrical coordinates?
(c) Which points P in R3 have the same rectangular
and spherical coordinates?
37. (a) How are the graphs of the polar equations r = f(9)
andr = -f{9) related?
(b) How are the graphs of the spherical equations
p = f((p, 9) and p = -f(q>, 0) related?
(c) Repeat part (a) for the graphs of r = f{9) and
r = 3/(0).
(d) Repeat part (b) for the graphs of p = f((p, 9) and
p = 3f(<p,0).
38. Suppose that a surface has an equation in cylindrical
coordinates of the form z = f(r). Explain why it must
be a surface of revolution.
39. (a) Verify that the basis vectors er, eg, and ez for
cylindrical coordinates are mutually perpendicu-
lar unit vectors.
(b) Verify that the basis vectors ep , , and eg for spher-
ical coordinates are mutually perpendicular unit
vectors.
40. Use the formulas in (8) to express i, j, k in terms of er,
eg , and ez .
41 . Use the formulas in (9) to express i, j, k in terms of ep,
ev, and eg.
42. Consider the solid in R3 shown in Figure 1.110.
(a) Describe the solid, using spherical coordinates.
(b) Describe the solid, using cylindrical coordinates.
A portion of the
z sphere of radius 3
(centered at origin)
y
x
Figure 1.110 The ice-cream-
cone-like solid in R3 in Exercise 42.
In Exercises 43-47, you will use the equations in (10) to es-
tablish those in (12).
43. Show that tan <pn-\ = x%lx\.
44. (a) Calculate x\ + x\ in terms of the hyperspherical
coordinates p, (pi, . . . , (pn-%.
(b) Assuming the inequalities in (1 1), use part (a) to
show that tamp„_2 = ^jx\ + x2/x^.
45. (a) Calculate x\ + x\ + x2 in terms of the hyperspher-
ical coordinates p, <p\, . .. , <p„-z.
(b) Assuming the inequalities in (1 1), use part (a) to
show that tan ip,,- 3 = J 1 x\ + x\ + x2/x4.
46. (a) For k — 2, . . . , n — 1, show that x\ + x\ + ■ ■ ■ +
x\ = p2 sin2 <p\ • • ■ sin2 (p„-k- (Note: This is best
accomplished by means of mathematical induc-
tion.)
(b) Assuming the inequalities in (11), use part (a)
to show that, for k = 2 n — 1, tan^„_j. =
jx2 + ---+xt/xk+l.
47. Show that x\ + x\ H \-x2 = p2.
Miscellaneous Exercises for Chapter 1
True/False Exercises for Chapter 1
1. If a = (1, 7, -9) and b = (1, -9, 7), then a = b.
2. If a and b are two vectors in R3 and k and / are real
numbers, then (k — Z)(a + b) = ka — /a + kb — lb.
3. The displacement vector from Pi (1,0, — 1) to
P2(5,3,2) is (-4,-3,-3).
4. Force and acceleration are vector quantities.
5. Velocity and speed are vector quantities.
6. Displacement and distance are scalar quantities.
7. If a particle is at the point (2, —1) in the plane and
moves from that point with velocity vector v = (1, 3),
then after 2 units of time have passed, the particle will
be at the point (5, 1).
8. The vector (2,3, —2) is the same as 2i + 3j — 2k.
9. A set of parametric equations for the line through
(1, -2, 0) that is parallel to (-2, 4, 7) is x = 1 - 2t,
y = 4t-2,z = l.
10. A set of parametric equations for the line through
(1, 2, 3) and (4, 3, 2) is x = 4 - 3f, y = 3 - t, z =
t + 2.
1 1 . The line with parametric equations x = 2 —
3t, y = t+l, z = 2t — 3 has symmetric form
x+2 _ _l_z-3
-3 ~y~ ~ 2
12. The two sets of parametric equations x = 3t — 1,
y = 2 - t, z = 2t + 5 and x = 2 - 6t, y = 2t + 1,
z = 1 — At both represent the same line.
13. The parametric equations x = 2sinf, y = 2cos?,
where 0 < t < n, describe a circle of radius 2.
1 4. The dot product of two unit vectors is 1 .
15. For any vector a in R" and scalar k, we have ||£a|| =
fc||a||.
16. If a, u e R" and ||u|| = 1, then projua = (a • u)u.
17. For any vectors a, b, c in R3, we have a x (b x c) =
(a x b) x c.
1 8. The volume of a parallelepiped determined by the vec-
tors a, b, c e R3 is |(a x c) • b|.
19. ||a||b — ||b||a is a vector.
20. (a x b) • c — (a x c) • b is a scalar.
21. The plane containing the points (1, 2, 1), (3, —1, 0),
and (1, 0, 2) has equation 5a: + 2y + 4z = 13.
22. The plane containing the points (1, 2, 1), (3, —1, 0),
and (1,0,2) is given by the parametric equations
x = 2s, y = —3s — 2t, z = t — s.
23. If A is a 5 x 7 matrix and B is a 7 x 7 matrix, then BA
is a 7 x 5 matrix.
24. If A
1 2 0
-10 2
5 9 2
0 8 0
then
-1
2
1
1
0
3
detA = 2
5
2
0
+ 9
-1
2
1
0
0
-6
0
0
-6
1
0
3
-8
1
2
1
5
2
0
25. If A is an n x n matrix, then det (2A) = 2 det A.
26. The surface having equation r = 4 sin 9 in cylindrical
coordinates is a cylinder of radius 2.
27. The surface having equation p = 4 cos 6 in spherical
coordinates is a sphere of radius 2.
28. The surface having equation p cos 6 sin cp = 3 in spher-
ical coordinates is a plane.
29. The surface having equation p = 3 in spherical coor-
dinates is the same as the surface whose equation in
cylindrical coordinates is r2 + z2 = 9.
30. The surface whose equation in cylindrical coordinates
is z = 2r is the same as the surface whose equation in
spherical coordinates is <p = jt/6.
Miscellaneous Exercises for Chapter 1
1 . If P\ , Pi, . . . , Pn are the vertices of a regular polygon
having n sides and if O is the center of the polygon,
show that Y^=\ OP, = 0. The case n = 5 is shown in
Figure 1.111. (Hint: Don't try using coordinates. Use
instead sketches, geometry, and perhaps translations or
rotations.)
Chapter 1 | Vectors
p4 P,
Figure 1.111 The case n = 5.
2. Find parametric equations for the line through the point
(1, 0, —2) that is parallel to the line x = 3t + 1, y =
5 - 7r, z = ; + 12.
3. Find parametric equations for the line through the
point (1, 0, —2) that intersects the line x = 3t + 1,
y = 5 — It , z = t + 12 orthogonally. (Hint: Let Xq =
3% + 1, yo = 5 — ltd, zo = to + 12 be the point where
the desired line intersects the given line.)
4. Given two points Po(ai, cti, ai) and P\(b\, £>2, fo), we
have seen in equations (3) and (4) of §1.2 how to
parametrize the line through Pq and Pi as r(r) =
O Pq + t Pq Pi , where t can be any real number. (Recall
that r = OP, the position vector of an arbitrary point
P on the line.)
(a) For what value of / does r(f) = OPb? For what
value of t does r(r) = OP\l
(b) Explain how to parametrize the line segment join-
ing Pq and Pi. (See Figure 1.112.)
z
y
Figure 1.112 The segment joining Po
and Pi is a portion of the line containing
Po and P|. (See Exercise 4.)
(c) Give a set of parametric equations for the line seg-
ment joining the points (0, 1, 3) and (2, 5, —7).
5. Recall that the perpendicular bisector of a line seg-
ment in R2 is the line through the midpoint of the seg-
ment that is orthogonal to the segment.
(a) Give a set of parametric equations for the perpen-
dicular bisector of the segment joining the points
Pi(-l,3)andP2(5, -7).
(b) Given general points Pi(ai,fl2) and P2(b\,b2),
provide a set of parametric equations for the per-
pendicular bisector of the segment joining them.
6. If we want to consider a perpendicular bisector of a
line segment in R3 , we will find that the bisector must
be a plane.
(a) Give an (implicit) equation for the plane that serves
as the perpendicular bisector of the segment join-
ing the points P(6, 3, -2) and P2(-4, 1, 0).
(b) Given general points P\{a\, 02, 03) and
Pi{b\, bj, ^3), provide an equation for the plane
that serves as the perpendicular bisector of the
segment joining them.
7. Generalizing Exercises 5 and 6, we may define the per-
pendicular bisector of a line segment in R" to be the
hyperplane through the midpoint of the segment that
is orthogonal to the segment.
(a) Give an equation for the hyperplane in R5 that
serves as the perpendicular bisector of the seg-
ment joining the points Pi(l, 6, 0, 3, — 2) and
P2(-3,-2,4, 1,0).
(b) Given arbitrary points P\(a\ , . . . , a„) and
Pzibi, , . . , bn) in R", provide an equation for
the hyperplane that serves as the perpendicular
bisector of the segment joining them.
8. If a and b are unit vectors in R3, show that
|| a x b||2 + (a-b)2 = 1.
9. (a) If a • b = a • c, does it follow that b = c? Explain
your answer.
(b) If a x b = a x c, does it follow that b = c?
Explain.
10. Show that the two lines
h : x = t-3, y=\-2t, z = 2t + 5
h : x = 4 - It, y = At + 3, z = 6 - At
are parallel, and find an equation for the plane that
contains them.
1 1 . Consider the two planes x + y = 1 and y + z = 1 .
These planes intersect in a straight line.
( a) Find the ( acute) angle of intersection between these
planes.
(b) Give a set of parametric equations for the line of
intersection.
1 2. Which of the following lines whose parametric equa-
tions are given below are parallel? Are any the same?
(a) x = At + 6, y = 2 - It, z = 8* + 1
Miscellaneous Exercises for Chapter 1
(b) x = 3 - 6t , y = 3f , z = 4 - 9t
(c) x = 2 - 2t, y = t + 4, z = -4r - 7
(d) jc = 2f + 4, y = 1 - f , z = 3? - 2
13. Determine which of the planes whose equations are
given below are parallel and which are perpendicular.
Are any of the planes the same?
(a) 2x + 3y - z = 3
(b) -6x + 4y - 2z + 2 = 0
(c) x + y - z = 2
(d) 10* + I5y - 5z = 1
(e) 3x - 2y + z = 1
14. (a) What is the angle between the diagonal of a cube
and one of the edges it meets? (Hint: Locate the
cube in space in a convenient way.)
(b) Find the angle between the diagonal of a cube and
the diagonal of one of its faces.
15. Mark each of the following statements with a 1 if you
agree, — 1 if you disagree:
(1) Red is my favorite color.
(2) I consider myself to be a good athlete.
(3) I like cats more than dogs.
(4) I enjoy spicy foods.
(5) Mathematics is my favorite subject.
Your responses to the preceding "questionnaire" may
be considered to form a vector in R5 . Suppose that you
and a friend calculate your respective "response vec-
tors" for the questionnaire. Explain the significance of
the dot product of your two vectors.
16. The median of a triangle is the line segment that joins
a vertex of a triangle to the midpoint of the opposite
side. The purpose of this problem is to use vectors to
show that the medians of a triangle all meet at a point.
(a) Using Figure 1.113, write the vectors B M\ and C Mi
in terms of AB and AC.
(b) Let P be the point of intersection of BM\ and CMi.
Write BP and CP in terms of AB and AC.
(c) Use the fact that CB = CP + Jb = CA + AB to
show that P must lie two-thirds of the way from B
to Mi and two-thirds of the way from C to Mi.
(d) Now use part (c) to show why all three medians must
meet at P .
17. Suppose that the four vectors a, b, c, and d in R3 are
coplanar (i.e., that they all lie in the same plane). Show
that then (a x b) x (c x d) = 0.
18. Show that the area of the triangle, two of whose
sides are determined by the vectors a and b (see
Figure 1.114), is given by the formula
Area = ^7||a||2||b||2 - (a • b)2.
Figure 1.1 14 The triangle in Exercise 18.
19. Let A(l,3,-1), 5(4,-1,3), C(2, 5,2), and
D(5, 1, 6) be the vertices of a parallelogram.
(a) Find the area of the parallelogram.
(b) Find the area of the projection of the parallelogram
in the xy -plane.
20. (a) For the line / in R2 given by the equation ax +
by = d, find a vector v that is parallel to /.
(b) Find a vector n that is normal to / and has first
component equal to a.
(c) If Po(-*o, Vo) is any point in R2, use vectors to de-
rive the following formula for the distance from
P0 to /:
\axo + bye — d\
Distance from Pq to / =
Figure 1.113 Two of the three medians
of a triangle in Exercise 16.
Va2 + b2
To do this, you'll find it helpful to use Figure 1.115,
where Pi(xu yi) is any point on I.
(d) Find the distance between the point (3, 5) and the
line 8jc — 5y = 2.
21. (a) If PoC^o, yo, Zo) is any point in R3, use vectors
to derive the following formula for the distance
from Po to the plane n having equation Ax +
By + Cz = D:
\Ax0 + By0 + Czo- D\
Distance from Pq to fl
VA2 + B2 + C2
78 Chapter 1 [ Vectors
y
Pa
\ n /
I: ax + by =
Figure 1.115 Geometric construction for
Exercise 20.
Figure 1.116 should help. (P\{x\, yi, Zi) is any
point in n.)
Distance
~-*P^J
Yl:Ax + By +
X
Figure 1.116 Geometric construction for
Exercise 2 1 .
(b) Find the distance between the point (1,5, —3) and
the plane x — 2y + 2z + 12 = 0.
22. (a) Let P be a point in space that is not contained in the
plane n that passes through the three noncollinear
points A , B , and C . Show that the distance between
P and n is given by the expression
|p-(bxc)|
lib x o ||
where p = AP, b = A§, and c = AC.
(b) Use the result of part (a) to find the distance
between (1, 0, —1) and the plane containing the
points (1, 2, 3), (2, -3, 1), and (2,-1, 0).
23. Let A, B, C, and D denote four distinct points in R3.
(a) Show that A, B, and C are collinear if and only if
Xi x A~t = 0.
(b) Show that A, B,C, and D are coplanar if and only
if (AT? x At)-ci> = Q.
24. Let x = OP, the position vector of a point P in R3.
Consider the equation
x-k 1
II || ~ V2'
Describe the configuration of points P that satisfy the
equation.
25. Let a and b be two fixed, nonzero vectors in R3 , and let
c be a fixed constant. Explain how the pair of equations,
a • x = c
a x x = b ,
completely determines the vector x e R3 .
26. (a) Give examples of vectors a, b, c in R3 that show
that, in general, it is not true that a x (b x c) =
(a x b) x c. (That is, the cross product is not as-
sociative.)
(b) Use the Jacobi identity (see Exercise 30 of §1.4)
to show that, for any vectors a, b, c in R3,
a x (b x c) = (a x b) x c
if and only if
(c x a) x b = 0.
27. (a) Given an arbitrary (i.e., not necessarily regular)
tetrahedron, associate to each of its four triangular
faces a vector outwardly normal to that face with
length equal to the area of that face. (See Fig-
ure 1.117.) Show that the sum of these four vec-
tors is zero. (Hint: Describe Vi , . . . , V4 in terms of
some of the vectors that run along the edges of the
tetrahedron.)
v3
Figure 1.117 The tetrahedron of part
(a) of Exercise 27.
(b) Recall that a polyhedron is a closed surface in
R3 consisting of a finite number of planar faces.
Suppose you are given the two tetrahedra shown
in Figure 1.118 and that face ABC of one is con-
gruent to face A' B'C' of the other. If you glue the
tetrahedra together along these congruent faces,
then the outer faces give you a six-faced polyhe-
dron. Associate to each face of this polyhedron an
outward-pointing normal vector with length equal
to the area of that face. Show that the sum of these
six vectors is zero.
Miscellaneous Exercises for Chapter 1
C
A
B B
Figure 1.118 In Exercise 27(b), glue the two tetrahedra shown along congruent faces.
28.
(c) Outline a proof of the following: Given an n-faced
polyhedron, associate to each face an outward-
pointing normal vector with length equal to the
area of that face. Show that the sum of these n
vectors is zero.
Consider a right tetrahedron, that is, a tetrahedron
that has a vertex R whose three adjacent faces are pair-
wise perpendicular. (See Figure 1.119.) Use the result
of Exercise 27 to show the following three-dimensional
analogue of the Pythagorean theorem: If a, b, and
c denote the areas of the three faces adjacent to R
and d denotes the area of the face opposite R, then
d2.
Figure 1.119 The right tetrahedron
of Exercise 28. The three faces
containing the vertex R are pairwise
perpendicular.
29. (a) Use vectors to prove that the sum of the squares
of the lengths of the diagonals of a parallelogram
equals the sum of the squares of the lengths of the
four sides.
(b) Give an algebraic generalization of part (a) for R" .
30. Show that for any real numbers a\
we have
n2
■ ■ , an,bi,
n
31 . To raise a square (n x n) matrix A to a positive integer
power n, one calculates A" as A ■ A ■ ■ ■ A (n times),
(a) Calculate successive powers A, A2, A3, A4 of the
1 1
matrix A = q \
(b) Conjecture the general form of A" for the matrix
A of part (a), where n is any positive integer.
(c) Prove your conjecture in part (b) using mathemat-
ical induction.
32. A square matrix A is called nilpotent if A"
some positive power n,
0 1 1
0 for
(a) Show that A
is nilpotent.
0 0 0
0 0 0
(b) Use a calculator or computer to show that A =
0 0 0 0 0
is nilpotent.
^ 33.
The n x n matrix Hn whose ijth entry is l/(i + j — 1)
is called the Hilbert matrix of order n .
(a) Write out H2, H], H^, H5, and H^, Use a com-
puter to calculate their determinants exactly. What
seems to happen to det Hn as n gets larger?
(b) Now calculate #10 and det H\q. If you use exact
arithmetic, you should find that det H\q / 0 and
hence that //10 is invertible. (See Exercises 30-38
of § 1 .6 for more about invertible matrices.)
(c) Now give a numerical approximation A for H\q.
Calculate the inverse matrix B of this approxima-
tion, if your computer allows. Then calculate AB
and B A . Do you obtain the 10x10 identity matrix
ha in both cases?
(d) Explain what parts (b) and (c) suggest about the
difficulties in using numerical approximations in
matrix arithmetic.
As a child, you may have played with a popular toy called a
Spirograph®. With it one could draw some appealing geomet-
ricfigures. The Spirograph consists of a small toothed disk with
several holes in it and a larger ring with teeth on both inside
and outside as shown in Figure 1.120. You can draw pictures by
meshing the small disk with either the inside or outside circles
of the ring and then poking a pen through one of the holes of
the disk while turning the disk. (The large ring is held fixed.)
80 Chapter 1 [ Vectors
An idealized version of the Spirograph can be obtained
by taking a large circle (of radius a) and letting a small circle
(of radius b) roll either inside or outside it without slipping.
A "Spirograph " pattern is produced by tracking a particu-
lar point lying anywhere on (or inside) the small circle. Exer-
cises 34-37 concern this set-up.
34. Suppose that the small circle rolls inside the larger
circle and that the point P we follow lies on the circum-
ference of the small circle. If the initial configuration
is such that P is at (a, 0), find parametric equations
for the curve traced by P, using angle t from the posi-
tive .r-axis to the center B of the moving circle. (This
configuration is shown in Figure 1.121.) The result-
ing curve is called a hypocycloid. Two examples are
shown in Figure 1.122.
y
a
(b. y\
fp ^\
At ( , b\
Figure 1.121 The coordinate
configuration for finding parametric
equations for a hypocycloid.
35. Now suppose that the small circle rolls on the outside
of the larger circle. Derive a set of parametric equa-
tions for the resulting curve in this case. Such a curve
is called an epicycloid, shown in Figure 1.123.
36. (a) A cusp (or corner) occurs on either the hypocy-
cloid or epicycloid every time the point P on the
small circle touches the large circle. Equivalently,
y
Figure 1.123 An epicycloid with
a = 4,b= 1.
this happens whenever the smaller circle rolls
through 2tt. Assuming that a/b is rational, how
many cusps does a hypocycloid or epicycloid
have? (Your answer should involve a and b in some
way.)
(b) Describe in words and pictures what happens when
a/b is not rational.
37. Consider the original Spirograph set-up again. If we
now mark a point P at a distance c from the center
of the smaller circle, then the curve traced by P is
called a hypotrochoid (if the smaller circle rolls on
the inside of the larger circle) or an epitrochoid (if
the smaller circle rolls on the outside). Note that we
must have b < a, but we can have c either larger or
smaller than b. (If c < b, we get a "true" Spirograph
pattern in the sense that the point P will be on the
inside of the smaller circle. The situation when c > b
is like having P mounted on the end of an elongated
spoke on the smaller circle.) Give a set of parametric
equations for the curves that result in this way. (See
Figure 1.124.)
Exercises 38-43 are made feasible through the use of appropri-
ate software for graphing in polar, cylindrical, and spherical
Miscellaneous Exercises for Chapter 1
coordinates. (Note: When using software for graphing in spher-
ical coordinates, be sure to check the definitions that are used
for the angles <p and 8.)
x
Figure 1.124 The configuration for finding
parametric equations for epitrochoids.
38. (a) Graph the curve in R2 whose polar equation is
r = cos 28.
(b) Graph the surface in R3 whose cylindrical equation
is r = cos 28.
(c) Graph the surface in R3 whose spherical equation
is p = cos 2<p.
(d) Graph the surface in R3 whose spherical equation
is p = cos 28.
39. (a) Graph the curve in R2 whose polar equation is
r = sin2#.
(b) Graph the surface in R3 whose cylindrical equation
is r = sin 2$.
(c) Graph the surface in R3 whose spherical equation
is p = sin2<p.
(d) Graph the surface in R3 whose spherical equation
is p = sin 28. Compare the results of this exercise
with those of Exercise 38.
40. (a) Graph the curve in R2 whose polar equation is
r = cos 38.
(b) Graph the surface in R3 whose cylindrical equation
is r = cos 38.
(c) Graph the surface in R3 whose spherical equation
is p = cos 3<p.
(d) Graph the surface in R3 whose spherical equation
is p = cos 38.
41. (a) Graph the curve in R2 whose polar equation is
r = sin 38.
(b) Graph the surface in R3 whose cylindrical equation
is r = sin 38.
(c) Graph the surface in R3 whose spherical equation
is p = sin 3(p.
(d) Graph the surface in R3 whose spherical equation
is p = sin 38. Compare the results of this exercise
with those of Exercise 40.
42. (a) Graph the curve in R2 whose polar equation is
r — 1 + sin |. (This curve is known as a nephroid,
meaning "kidney shaped.")
(b) Graph the surface in R3 whose cylindrical equation
is r = 1 + sin ~ .
(c) Graph the surface in R3 whose spherical equation
is p = 1 + sin | .
(d) Graph the surface in R3 whose spherical equation
is p = 1 + sin | .
43. (a) Graph the curve in R2 whose polar equation is
r = 8.
(b) Graph the surface in R3 whose cylindrical equation
is r = 8.
(c) Graph the surface in R3 whose spherical equation
is p = <p,
(d) Graph the surface in R3 whose spherical equation
is p = 8, where jt/2 < cp < n and 0 < 8 < 4n.
44. Consider the solid hemisphere of radius 5 pictured in
Figure 1.125.
(a) Describe this solid, using spherical coordinates.
(b) Describe this solid, using cylindrical coordinates.
z
y
Figure 1.125 The solid hemisphere of
Exercise 44.
45. Consider the solid cylinder pictured in Figure 1.126.
(a) Describe this solid, using cylindrical coordinates
(position the cylinder conveniently).
(b) Describe this solid, using spherical coordinates.
h 6 -
Figure 1.126 The solid
cylinder of Exercise 45.
2.1 Functions of Several
Variables; Graphing
Surfaces
2.2 Limits
2.3 The Derivative
2.4 Properties; Higher-order
Partial Derivatives
2.5 The Chain Rule
2.6 Directional Derivatives and
the Gradient
2.7 Newton's Method (optional)
True/False Exercises for
Chapter 2
Miscellaneous Exercises for
Chapter 2
/
X Y
Figure 2.1 The mapping
nature of a function.
Differentiation in
Several Variables
2.1 Functions of Several Variables;
Graphing Surfaces
The volume and surface area of a sphere depend on its radius, the formulas
describing their relationships being V = |:rr3 and S = 4nr2. (Here V and S
are, respectively, the volume and surface area of the sphere and r its radius.)
These equations define the volume and surface area as functions of the radius.
The essential characteristic of a function is that the so-called independent variable
(in this case the radius) determines a unique value of the dependent variable ( V
or S). No doubt you can think of many quantities that are determined uniquely
not by one variable (as the volume of a sphere is determined by its radius) but
by several: the area of a rectangle, the volume of a cylinder or cone, the average
annual rainfall in Cleveland or the national debt. Realistic modeling of the world
requires that we understand the concept of a function of more than one variable
and how to find meaningful ways to visualize such functions.
Definitions, Notation, and Examples
A function, any function, has three features: (1) a domain set X, (2) a codomain
set Y, and (3) a rule of assignment that associates to each element x in the
domain X a unique element, usually denoted f(x), in the codomain Y. We will
frequently use the notation f:X—> Y for a function. Such notation indicates all
the ingredients of a particular function, although it does not make the nature of
the rule of assignment explicit. This notation also suggests the "mapping" nature
of a function, indicated by Figure 2.1.
EXAMPLE 1 Abstract definitions are necessary, but it is just as important that
you understand functions as they actually occur. Consider the act of assigning to
each U.S. citizen his or her social security number. This pairing defines a function:
Each citizen is assigned one social security number. The domain is the set of U.S.
citizens and the codomain is the set of all nine-digit strings of numbers.
On the other hand when a university assigns students to dormitory rooms, it
is unlikely that it is creating a function from the set of available rooms to the set of
students. This is because some rooms may have more than one student assigned
to them, so that a particular room does not necessarily determine a unique student
occupant. ♦
2.1 | Functions of Several Variables; Graphing Surfaces
DEFINITION 1 .1 The range of a function f:X ^ Y is the set of those
elements of Y that are actual values of /. That is, the range of / consists of
those y in Y such that y = f(x) for some x in X.
Using set notation, we find that
Range / = {y e Y \ y = f(x) for some x e X] .
In the social security function of Example 1, the range consists of those nine-
digit numbers actually used as social security numbers. For example, the number
000-00-0000 is not in the range, since no one is actually assigned this number.
Figure 2.2 Every y e Y is "hit"
by at least one x e X.
Figure 2.3 The element b e Y
is not the image of any x £ X.
DEFINITION 1 .2 A function /: X -> Y is said to be onto (or surjective)
if every element of Y is the image of some element of X, that is, if range
f=Y.
The social security function is not onto, since 000-00-0000 is in the codomain
but not in the range. Pictorially, an onto function is suggested by Figure 2.2. A
function that is not onto looks instead like Figure 2.3. You may find it helpful to
think of the codomain of a function / as the set of possible (or allowable) values
of /, and the range of / as the set of actual values attained. Then an onto function
is one whose possible and actual values are the same.
DEFINITION 1 .3 A function /: X Y is called one-one (or injective) if
no two distinct elements of the domain have the same image under /. That
is, / is one-one if whenever x\, *2 £ X and x\ ^ x2, then f{x\) ^ f(xz).
(See Figure 2.4.)
not one-one
Figure 2.4 The figure on the left depicts a one-one mapping; the one
on the right shows a function that is not one-one.
One would expect the social security function to be one-one, but we have heard
of cases of two people being assigned the same number so that, alas, apparently
it is not.
When you studied single-variable calculus, the functions of interest were
those whose domains and codomains were subsets of R (the real numbers). It
was probably the case that only the rule of assignment was made explicit; it is
generally assumed that the domain is the largest possible subset of R for which
the function makes sense. The codomain is generally taken to be all of R.
84 Chapter 2 I Differentiation in Several Variables
EXAMPLE 2 Suppose /: R —> R is given by /(jc) = x2. Then the domain and
codomain are, explicitly, all of R, but the range of / is the interval [0, oo). Thus
/ is not onto, since the codomain is strictly larger than the range. Note that / is
not one-one, since f(2) = /(— 2) = 4, but 2 —2. ♦
EXAMPLE 3 Suppose g is a function such that g(x) = >Jx — 1. Then if we
take the codomain to be all of R, the domain cannot be any larger than [1, oo).
If the domain included any values less than one, the radicand would be negative
and, hence, g would not be real-valued. ♦
Now we're ready to think about functions of more than one real variable. In
the most general terms, these are the functions whose domains are subsets X of
R" and whose codomains are subsets of R"', for some positive integers n and m.
(For simplicity of notation, we'll take the codomains to be all of R'", except when
specified otherwise.) That is, such a function is a mapping f:XQ R" R™ that
associates to a vector (or point) x in X a unique vector (point) f(x) in R"! .
EXAMPLE 4 Let T: R3 -* R be defined by T(x, y, z) = xy + xz + yz. We
can think of T as a sort of "temperature function." Given a point x = (x, y, z) in
R3, T(x) calculates the temperature at that point. ♦
EXAMPLE 5 Let L: R" -> R be given by L(x) = ||x|| . This is a "length func-
tion" in that it computes the length of any vector x in R". Note that L is not
one-one, since L(e,-) = L(ey) = 1, where e, and e;- are any two of the standard
basis vectors for R" . L also fails to be onto, since the length of a vector is always
nonnegative. ♦
EXAMPLE 6 Consider the function given by N(x) = x/||x|| where x is a vector
in R3 . Note that N is not defined if x = 0, so the largest possible domain for N is
R3 — {0}. The range of N consists of all unit vectors in R3. The function N is the
"normalization function," that is, the function that takes a nonzero vector in R3
and returns the unit vector that points in the same direction. ♦
EXAMPLE 7 Sometimes a function may be given numerically by a table.
One such example is the notion of windchill — the apparent temperature one
feels when taking into account both the actual air temperature and the speed of
the wind. A standard table of windchill values is shown in Figure 2.5. 1 From it
we see that if the air temperature is 20 °F and the windspeed is 25 mph, the wind-
chill temperature ("how cold it feels") is 3 °F. Similarly, if the air temperature is
35°F and the windspeed is 10 mph, then the windchill is 27°F. In other words,
if s denotes windspeed and t air temperature, then the windchill is a function
W(s, t). ♦
The functions described in Examples 4, 5, and 7 are scalar-valued functions,
that is, functions whose codomains are R or subsets of R. Scalar- valued functions
are our main concern for this chapter. Nonetheless, let's look at a few examples
of functions whose codomains are R"! where m > 1 .
NOAA, National Weather Service, Office of Climate, Water, and Weather Services, "NWS Wind Chill
Temperature Index." February 26, 2004. <http://www.nws.noaa.gov/om/windchill> (July 31, 2010).
2.1 | Functions of Several Variables; Graphing Surfaces
Air Temp
(deg F)
Windspeed (mph)
5
10
15
20
25
30
35 40
45
50
55
60
40
36
34
32
30
29
28
28 27
26
26
25
25
35
31
27
25
24
23
22
21 20
19
19
18
17
30
25
21
19
17
16
15
14 13
12
12
11
10
25
19
15
13
11
9
8
7 6
5
4
4
3
20
13
9
6
4
3
1
0 -1
-2
-3
-3
-4
15
7
3
0
-2
-4
-5
-7 -8
-9
-10
-11
-11
10
1
-4
-7
-9
-11
-12
-14 -15
-16
-17
-18
-19
5
-5
-10
-13
-15
-17
-19
-21 -22
-23
-24
-25
-26
0
-11
-16
-19
-22
-24
-26
-27 -29
-30
-31
-32
-33
-5
-16
-22
-26
-29
-31
-33
-34 -36
-37
-38
-39
-40
-10
-22
-28
-32
-35
-37
-39
-41 -43
-44
-45
-46
-48
-15
-28
-35
-39
-42
-44
-46
-48 -50
-51
-52
-54
-55
-20
-34
-41
-45
-48
-51
-53
-55 -57
-58
-60
-61
-62
-25
-40
-47
-51
-55
-58
-60
-62 -64
-65
-67
-68
-69
-30
-46
-53
-58
-61
-64
-67
-69 -71
-72
-74
-75
-76
-35
-52
-59
-64
-68
-71
-73
-76 -78
-79
-81
-82
-84
-40
-57
-66
-71
-74
-78
-80
-82 -84
-86
-88
-89
-91
-45
-63
-72
-77
-81
-84
-87
-89 -91
-93
-95
-97
-98
Figure 2.6 The helix of
Example 8. The arrow shows the
direction of increasing t .
Figure 2.5 Table of windchill values in English units.
EXAMPLE 8 Define f: R -> R3 by f(f) = (cos t, sin?, t). The range off is the
curve in R3 with parametric equations x = cos t , y = sin t , z = t . If we think of
t as a time parameter, then this function traces out the corkscrew curve (called a
helix) shown in Figure 2.6. ♦
EXAMPLE 9 We can think of the velocity of a fluid as a vector in R3. This
vector depends on (at least) the point at which one measures the velocity and also
the time at which one makes the measurement. In other words, velocity may be
considered to be a function v: X c R4 — >• R3. The domain X is a subset of R4
because three variables x, y, z are required to describe a point in the fluid and a
fourth variable t is needed to keep track of time. (See Figure 2.7.) For instance,
such a function v might be given by the expression
v(x, y, z, t) = xyzt\ + (x1
y2)l
(3z + /)k.
Figure 2.7 A water
pitcher. The velocity v of
the water is a function
from a subset of R4 to R3.
You may have noted that the expression for v in Example 9 is considerably
more complicated than those for the functions given in Examples 4-8. This is
because all the variables and vector components have been written out explicitly.
In general, if we have a function f: X c R" — > Rm, then xeX can be written as
x = (xi , X2, ■ ■ ■ , x„) and f can be written in terms of its component functions
/i. fz< ■ ■ ■ . fm- The component functions are scalar- valued functions of x £ X
that define the components of the vector f(x) £ R"! . What results is a morass of
symbols:
f(x) = f(jti, X2, ■ . ■ , xn) (emphasizing the variables)
— (/i(x)> /2(x)> ■ • • > /m(x)) (emphasizing the component functions)
= {fi{x\,x2, . . .,xn), f2(xi,x2, . . . ,Xn), fm(Xl,X2, ■ ■ .,x„))
(writing out all components).
86 Chapter 2 I Differentiation in Several Variables
For example, the function L of Example 5, when expanded, becomes
L(x) = L(x\,x2, . . . , x„) = ^Jxf + x\
The function N of Example 6 becomes
X (X],X2,X-i)
N(x) = —
^ llxll
x2 x3
and, hence, the three component functions of N are
X\ X2
Ni(xi,x2,X3) = N2(xi, x2, xi) =
X^ ~\~ X2 ~\~ X^ J X\ -\- x2 ~\~ x^
x3
Ni(xi,X2, x3) =
X\ -\- Xj ~"T~ X-i
Although writing a function in terms of all its variables and components has
the advantage of being explicit, quite a lot of paper and ink are used in the process.
The use of vector notation not only saves space and trees but also helps to make
the meaning of a function clear by emphasizing that a function maps points in
R" to points in R'" . Vector notation makes a function of 300 variables look "just
like" a function of one variable. Try to avoid writing out components as much as
you can (except when you want to impress your friends).
Visualizing Functions
No doubt you have been graphing scalar-valued functions of one variable for so
long that you give the matter little thought. Let's scrutinize what you've been do-
ing, however. A function /:XcR^ R takes a real number and returns another
real number as suggested by Figure 2.8. The graph of / is something that "lives"
in R2. (See Figure 2.9.) It consists of points (x, y) such that y = f(x). That is,
Graph / = {{x, f(x)) \ x e X} = {(x, y)\xeX,y = f(x)} .
The important fact is that, in general, the graph of a scalar-valued function
of a single variable is a curve — a one-dimensional object — sitting inside two-
dimensional space.
y
f
SKA
(x,f(x)) /
X
Figure 2.8 A function /:ICR->R, Figure 2.9 The graph off.
2.1 | Functions of Several Variables; Graphing Surfaces
Now suppose we have a function /:XcR2->R, that is, a function of two
variables. We make essentially the same definition for the graph:
Graph / = {(x, /(x)) xel). (1)
Of course, x = (x, y) is a point of R2. Thus, {(x, fix))} may also be written as
l(x, y, fix, y))} , or as {(x, y, z) | (x, y) e X, z = f(x, y)} .
Hence, the graph of a scalar-valued function of two variables is something that
sits in R3. Generally speaking, the graph will be a surface.
EXAMPLE 1 0 The graph of the function
9 1,1,7
/:R2^R /(x,y)=-y3-y--x2 + -
is shown in Figure 2.10. For each point x = (x, y) in R2, the point in R3 with
coordinates (x, y, ^y3 — y — \x2 + |) is graphed. ♦
Figure 2.10 The graph of fix, y) = -j^y3 — y — \x2 + |.
Graphing functions of two variables is a much more difficult task than graph-
ing functions of one variable. Of course, one method is to let a computer do
the work. Nonetheless, if you want to get a feeling for functions of more than
one variable, being able to sketch a rough graph by hand is still a valuable skill.
The trick to putting together a reasonable graph is to find a way to cut down on
the dimensions involved. One way this can be achieved is by drawing certain
special curves that lie on the surface z = fix, y). These special curves, called
contour curves, are the ones obtained by intersecting the surface with horizontal
planes z = c for various values of the constant c. Some contour curves drawn
on the surface of Example 10 are shown in Figure 2.1 1. If we compress all the
contour curves onto the xy-plane (in essence, if we look down along the posi-
tive z-axis), then we create a "topographic map" of the surface that is shown in
Figure 2.12. These curves in the xy-plane are called the level curves of the original
function /.
The point of the preceding discussion is that we can reverse the process in
order to sketch systematically the graph of a function / of two variables: We
88 Chapter 2 I Differentiation in Several Variables
Figure 2.1 1 Some contour curves of the Figure 2.1 2 Some level curves of
function in Example 10. the function in Example 10.
first construct a topographic map in R2 by finding the level curves of /, then
situate these curves in R3 as contour curves at the appropriate heights, and finally
complete the graph of the function. Before we give an example, let's restate our
terminology with greater precision.
DEFINITION 1 .4 Let /:XcR2^Rbea scalar-valued function of two
variables. The level curve at height c of / is the curve in R2 defined by the
equation f(x, y) = c, where c is a constant. In mathematical notation,
Level curve at height c = {(x, y) £ R2 | fix, y) = c] .
The contour curve at height c of / is the curve in R3 defined by the two
equations z = fix, y) and z = c. Symbolized,
Contour curve at height c = {(x, y, z) £ R3 | z = fix, y) = c} .
In addition to level and contour curves, consideration of the sections of a
surface by the planes where x or y is held constant is also helpful. A section of a
surface by a plane is just the intersection of the surface with that plane. Formally,
we have the following definition:
DEFINITION 1 .5 Let /: X C R2 -> R be a scalar-valued function of two
variables. The section of the graph of / by the plane x = c (where c
is a constant) is the set of points (x, y, z), where z = f(x, y) and x = c.
Symbolized,
Section by x = c is {(x, y, z) e R3 I z = f(x, y), x = c}.
Similarly, the section of the graph of / by the plane y = c is the set of
points described as follows:
Section by y = c is {(x, y, z) e R3 | z = fix, y), y = c].
2.1 | Functions of Several Variables; Graphing Surfaces
EXAMPLE 1 1 We'll use level and contour curves to construct the graph of the
function
/:R2^R, f(x,y) = 4-x2-y2.
By Definition 1 .4, the level curve at height c is
{(x, y) 6 R2 | 4 - x2 - y2 = c} = {(x, y) | x2 + y2 = 4 - c} .
Thus, we see that the level curves for c < 4 are circles centered at the origin of
radius -J4 — c. The level "curve" at height c = 4 is not a curve at all but just a
single point (the origin). Finally, there are no level curves at heights larger than 4
since the equation x2 + y2 = 4 — c has no real solutions in x and y. (Why not?)
These remarks are summarized in the following table:
c
Level curve x2 + y2 = 4 — c
-5
x2 + y2 = 9
-1
x2 + y2 = 5
0
x2 + y2 = 4
1
x2 + y2 = 3
3
x2 + y2 = l
4
x2 + y2 = 0 x = y = 0
c, where c > 4
empty
Thus, the family of level curves, the "topographic map" of the surface z =
4 — x2 — y2, is shown in Figure 2.13. Some contour curves, which sit in R3,
are shown in Figure 2.14, where we can get a feeling for the complete graph of
z = 4 — x2 — y2. It is a surface that looks like an inverted dish and is called a
paraboloid. (See Figure 2. 15.) To make the picture clearer, we have also sketched
in the sections of the surface by the planes x = 0 and y = 0. The section by x = 0
is given analytically by the set
{(x, y, z) e R3 | z = 4 - x2 - y2, x = 0} = {(0, y,z)\z = 4- y2} .
Similarly, the section by y = 0 is
{(x, y, z) e R3 | z = 4 - x2 - y2, y = 0} = {(x, 0, z) \ z = 4 - x2} .
Figure 2.13 The topographic Figure 2.14 Some contour Figure 2.15 The graph of
map of z = 4 — x2 — y2 (i.e., curves of z = 4 — x2 — y2. f(x, y) = 4 — x2 — y2 .
several of its level curves).
Chapter 2 | Differentiation in Several Variables
Since these sections are parabolas, it is easy to see how this surface obtained its
name. ♦
EXAMPLE 12 We'll graph the function g: R2 -> R, g(x, y) = y2 - x2. The
level curves are all hyperbolas, with the exception of the level curve at height 0,
which is a pair of intersecting lines.
c = -4
c
Level curve y2 — x2 = c
-4
x2 - y2 = 4
4
-1
x2 - y2 = 1
0
y2-X2=0
<S=> (y-x)(y+x) = 0 J=
=>■ y = ±x
X
1
y2-x2 = \
4
y2-X2=4
Figure 2.1 6 Some level curves
of g(x,y) = y2 -x2.
Figure 2.1 7 The contour curves
and graph of g(x, y) = y2 — x2.
The collection of level curves is graphed in Figure 2.16. The sections by x = c
are
l(x, y, z) | z = y2 - x2,x = c} = {(c, y, z) \ z = y2 - c2}.
These are clearly parabolas in the planes x = c. The sections by y = c are
{(x, y, z) | z = y2 - x2, y = c) = {(c, y, z) | z = c2 - x2},
which are again parabolas. The level curves and sections generate the contour
curves and surface depicted in Figure 2.17. Perhaps understandably, this surface
is called a hyperbolic paraboloid. ♦
EXAMPLE 13 We compare the graphs of the function f(x, y) = 4 — x2 — y2
of Example 1 1 with that of
h: R2 - {(0, 0)} -> R, h(x, y) = ln(x2 + y2).
The level curve of h at height c is
{(jc, y) 6 R2 | ln(x2 + y2) = c] = {(x, y) | x2 + y2 = ec\ .
Since ec > 0 for all c 6 R, we see that the level curve exists for any c and is a
circle of radius J~e^ = ec/2.
c
Level curve x2 + y2 = ec
-5
x2 + y2 = e~5
-1
x2 + y2 = e~l
0
x 2 + y2 = 1
1
x2 + y2 = e
3
x2 + y2 = e3
4
x2+y2 = e4
The collection of level curves is shown in Figure 2.18 and the graph in Figure 2.19.
Note that the section of the graph by x = 0 is
{(x,y,z) eR3 | Z = \n(x2 + y2),x = 0} = {(0, y, z) \ z = ln(y2) = 2 In | y | } .
2.1 | Functions of Several Variables; Graphing Surfaces
y
X
Figure 2.1 8 The collection of level curves ofz = ln(x2 + v2).
The section by _y = 0 is entirely similar:
{(x,y,z)eR3 | z = ln(x2 + y2), y = 0} = {(*,(>, z)| z = ln(x2) = 2 In \x\ } .
♦
In fact, if we switch from Cartesian to cylindrical coordinates, it is quite
easy to understand the surfaces in both Examples 11 and 13. In view of the
Cartesian/cylindrical relation x2 + y2 = r2, we see that for the function / of
Example 11,
z = 4 - x2 - y2 = 4 - (x2 + y2) = 4 - r2.
For the function h of Example 13, we have
z = ln(x2 + y2) = ln(r2) = 21nr,
where we assume the usual convention that the cylindrical coordinate r is non-
negative. Thus both of the graphs in Figures 2.15 and 2.19 are of surfaces of
revolution obtained by revolving different curves about the z-axis. As a result,
the level curves are, in general, circular.
The preceding discussion has been devoted entirely to graphing scalar- valued
functions of just two variables. However, all the ideas can be extended to more
variables and higher dimensions. If /: X c R" —> R is a (scalar-valued) function
of n variables, then the graph of / is the subset of R"+1 given by
Graph / = {(x, /(x)) xel}
= {{x\, ■ ■ ■ , X„, Xn+\)
(Xi, . . .,Xn) € X,
(2)
xn+\ = f(xl , ■ ■ ■ , Xn)} .
92 Chapter 2 | Differentiation in Several Variables
z
y
Figure 2.20 The level sets of the
function F(x, y,z) = x + y + z
are planes in R3 .
y
(1,0)
i x
Figure 2.21 The unit circle
x2 + y2 = 1.
Z
Figure 2.22 The sphere of radius
a, centered at (xq, yo, zo).
(The compactness of vector notation makes the definition of the graph of a function
of n variables exactly the same as in ( 1 ).) The level set at height c of such a function
is defined by
Level set at height c = {x e R" | f(x) = c}
= {(*i, X2, . . . , X„) | f(Xl, X2, ...,X») = C}.
While the graph of / is a subset of R"+1, a level set of / is a subset of R". This
makes it possible to get some geometric insight into graphs of functions of three
variables, even though we cannot actually visualize them.
EXAMPLE 14 Let F: R3 -> R be given by F(x, y, z) = x + y + z. Then the
graph of F is the set {(*, y, z, if) | w = x + y + z} and is a subset (called a
hypersurface) of R4, which we cannot depict adequately. Nonetheless, we can
look at the level sets of F, which are surfaces in R3. (See Figure 2.20.) We have
Level set at height c = {(x, y, z) \ x + y + z = c}.
Thus, the level sets form a family of parallel planes with normal vector i + j + k.
♦
Surfaces in General
Not all curves in R2 can be described as the graph of a single function of one
variable. Perhaps the most familiar example is the unit circle shown in Figure 2.2 1 .
Its graph cannot be determined by a single equation of the form y = f(x) (or, for
that matter, by one of the form x = g(y)). As we know, the graph of the circle may
be described analytically by the equation x2 + y2 = 1. In general, a curve in R2
is determined by an arbitrary equation in x and y, not necessarily one that isolates
y alone on one side. In other words, this means that a general curve is given by an
equation of the form F(x, y) — c (i.e., a level set of a function of two variables).
The analogous situation occurs with surfaces in R3 . Frequently a surface is
determined by an equation of the form F(x, y, z) — c (i.e., as a level set of a
function of three variables), not necessarily one of the form z = f(x, y).
EXAMPLE 15 A sphere is a surface in R3 whose points are all equidistant from
a fixed point. If this fixed point is the origin, then the equation for the sphere is
llx-OH = |WI =o, (3)
where a is a positive constant and x = (x, y, z) is a point on the sphere. If we
square both sides of equation (3) and expand the (implicit) dot product, then we
obtain perhaps the familiar equation of a sphere of radius a centered at the origin:
x2 + y2 + Z2 = a2. (4)
If the center of the sphere is at the point xo = (jto, yo, zq), rather than the origin,
then equation (3) should be modified to
l|x-x0||=a. (5)
(See Figure 2.22.)
When equation (5) is expanded, the following general equation for a sphere
is obtained:
(x - x0)2 + (y- y0)2 + (z - zo)2 = a2 (6)
2.1 | Functions of Several Variables; Graphing Surfaces
In the equation for a sphere, there is no way to solve for z uniquely in terms
of x and y. Indeed, if we try to isolate z in equation (4), then
2 2 2 2
z = a — x — y ,
so we are forced to make a choice of positive or negative square roots in order to
solve for z:
z = \J a1 — x2 — y2 or z. = —\J a2 — x2 — y2.
The positive square root corresponds to the upper hemisphere and the negative
square root to the lower one. In any case, the entire sphere cannot be the graph
of a single function of two variables. ♦
Of course, the graph of a function of two variables does describe a surface in
the "level set" sense. If a surface happens to be given by an equation of the form
z = f(x, y)
for some appropriate function /:XcR2^R, then we can move z to the oppo-
site side, obtaining
f{x, y)-z = 0.
If we define a new function F of three variables by
F{x,y,z) = f(x,y)-z,
then the graph of / is precisely the level set at height 0 of F. We reiterate this
point since it is all too often forgotten: The graph of a function of two variables is
a surface in R3 and is a level set of a function of three variables. However, not all
level sets of functions of three variables are graphs of functions of two variables.
We urge you to understand this distinction.
Quadric Surfaces
Conic sections, those curves obtained from the intersection of a cone with various
planes, are among the simplest, yet also the most interesting, of plane curves:
They are the circle, the ellipse, the parabola, and the hyperbola. Besides being
produced in a similar geometric manner, conic sections have an elegant algebraic
connection: Every conic section is described analytically by a polynomial equation
of degree two in two variables. That is, every conic can be described by an equation
that looks like
Ax2 + Bxy + Cy2 + Dx + Ey + F = 0
for suitable constants A, . . . , F.
In R3, the analytic analogue of the conic section is called a quadric surface.
Quadric surfaces are those defined by equations that are polynomials of degree
two in three variables:
Ax2 + Bxy + Cxz + Dy2 + Eyz + Fz2 + Gx + Hy + lz + J = 0.
To pass from this equation to the appropriate graph is, in general, a cumbersome
process without the aid of either a computer or more linear algebra than we
currently have at our disposal. So, instead we offer examples of those quadric
surfaces whose axes of symmetry lie along the coordinate axes in R3 and whose
corresponding analytic equations are relatively simple. In the discussion that
follows, a, b, and c are constants, which, for convenience, we take to be positive.
94 Chapter 2 | Differentiation in Several Variables
Z
Figure 2.26 The elliptic
cone
Figure 2.27 The graph of the
2 2 2
x y z
equation — r + -pr T = 1 is a
a2- bL cL
hyperboloid of one sheet.
Ellipsoid (Figure 2.23.) x2/a2 + y2/b2 + z2/c2 = 1.
This is the three-dimensional analogue of an ellipse in the plane. The sections
of the ellipsoid by planes perpendicular to the coordinate axes are all ellipses.
For example, if the ellipsoid is intersected with the plane z = 0, one obtains the
standard ellipse x2/a2 + y2/b2 = 1, z = 0. If a = b = c, then the ellipsoid is a
sphere of radius a.
Elliptic paraboloid (Figure 2.24.) z/c = x2/a2 + y2/b2.
(The roles of x, y, and z may be interchanged.) This surface is the graph of
a function of x and y. The paraboloid has elliptical (or single-point or empty)
sections by the planes "z = constant" and parabolic sections by "x = constant" or
" v = constant" planes. The constants a and b affect the aspect ratio of the elliptical
cross sections, and the constant c affects the steepness of the dish. (Larger values
of c produce steeper paraboloids.)
Hyperbolic paraboloid (Figure 2.25.) z/c = y2/b2 — x2/a2.
(Again the roles of x, y, and z may be interchanged.) We saw the graph of this
surface earlier in Example 12 of this section. It is shaped like a saddle whose "x =
constant" or "y = constant" sections are parabolas and "z = constant" sections
are hyperbolas.
Elliptic cone (Figure 2.26.) z2/c2 = x2/a2 + y2/b2.
The sections by "z = constant" planes are ellipses. The sections by x = 0 or
y = 0 are each a pair of intersecting lines.
Hyperboloid of one sheet (Figure 2.27.) x2/a2 + y2/b2 - z2/c2 = 1.
The term "one sheet" signifies that the surface is connected (i.e., that you can
travel between any two points on the surface without having to leave the surface).
The sections by "z = constant" planes are ellipses and those by "x = constant"
or "y = constant" planes are hyperbolas, hence, this surface's name.
Hyperboloid of two sheets (Figure 2.28.) z2/c2 - x2/a2 - y2/b2 = 1.
The fact that the left-hand side of the defining equation is the opposite of the left
side of the equation for the previous hyperboloid is what causes this surface to
consist of two pieces instead of one. More precisely, consider the sections of the
2.1 | Exercises 95
(0, 0, c)
Figure 2.28 The graph of
the equation
2 2 2
z x y
cL or bL
hyperboloid of two sheets.
Figure 2.29 The
hyperboloids
asymptotic to the cone
c2~ a2 + b2'
surface by planes of the form z = k for different constants k. These sections are
thus given by
or, equivalently, by
7i
2 n
X V
I- —
a2 b2
r
b2
e
= 1,
z = k
1, z = k.
If —c < k < c, then 0 < k2/c2 < 1. Thus, /:2/c2 — 1 < 0, and so the preceding
equation has no solution in x and y. Hence, the section by z = k, where \k\ < c,
is empty. If \k\ > c, then the section is an ellipse. The sections by "x = constant"
or "y = constant" planes are hyperbolas.
In the same way that the hyperbolas
2 n
x y
a2 b2
are asymptotic to the lines y = ±{b/a)x, the hyperboloids
= ±1
2 2 1
Z _ x yL
c2 a2 b2
± 1
are asymptotic to the cone
= X-, + y
b2'
This is perhaps intuitively clear from Figure 2.29, but let's see how to prove it
rigorously. In our present context, to say that the hyperboloids are asymptotic
to the cones means that they look more and more like the cones as |z| becomes
(arbitrarily) large. Analytically, this should mean that the equations for the hy-
perboloids should approximate the equation for the cone for sufficiently large \z\.
The equations of the hyperboloids can be written as follows:
r2 v2 72 72 / r1
x . y * ±1 * (i±*
a2 b2 cA c- \ zA
As \z\ -> oo, c2/z2 -> 0, so the right side of the equation for the hyperboloids
approaches z2/c2. Hence, the equations for the hyperboloids approximate that of
the cone, as desired.
2.1 Exercises
1. Let/:R^ R be given by f(x) = 2x2 + 1.
(a) Find the domain and range of /.
(b) Is / one-one?
(c) Is / onto?
2. Let g: R2 -> Rbe given by g(x, y) = 2x2 + 3y2 — 7.
(a) Find the domain and range of g.
(b) Find a way to restrict the domain to make a new
function with the same rule of assignment as g that
is one-one.
(c) Find a way to restrict the codomain to make a new
function with the same rule of assignment as g that
is onto.
Find the domain and range of each of the functions given in
Exercises 3-7.
3. fix, y)=-
y
4. f(x, y) = ln(x + y)
5. g(x, y, z) = Jx2 + (y- 2)2 + (z + I)2
Chapter 2 | Differentiation in Several Variables
6. g(x, y, z) =
7. f(x,y)= [x + y
y-1
x2+y2
8. Let f:R2 -> R3 be defined by f(x, y) = (x + y,
yex, x2y + 7). Determine the component functions
off.
9. Determine the component functions of the function v
in Example 9.
1 0. Let f : R3 -> R3 be defined by f(x) = x + 3 j . Write out
the component functions of f in terms of the compo-
nents of the vector x.
1 1 . Consider the mapping that assigns to a nonzero vector
x in R3 the vector of length 2 that points in the direction
opposite to x.
(a) Give an analytic (symbolic) description of this
mapping.
(b) If x = (x, y, z), determine the component func-
tions of this mapping.
1 2. Consider the function f: R2 —> R3 given by f(x) = Ax,
2 -1 "
where A = 5 0
-6 3
and the vector x in R is
written as the 2x1 column matrix x
X]
X2
(a) Explicitly determine the component functions of
f in terms of the components X\, X2 of the vector
(i.e., column matrix) x.
(b) Describe the range of f.
1 3. Consider the function f: R4 — > R3 given by f(x) = Ax,
~ 2 0 -1 1
0 3 0 0
2 0-11
and the vector x in R4
is written as the 4x1 column matrix x
x\
Xl
-V3
X4
(a) Determine the component functions of f in terms
of the components xi , xi, *3, X4 of the vector (i.e.,
column matrix) x.
(b) Describe the range of f.
In each of Exercises 14-23, (a) determine several level curves
of the given function f (make sure to indicate the height c of
each curve); (b) use the information obtained in part (a) to
sketch the graph of f.
14. f(x, y) = 3
15. f(x,y)
16. f(x,y) = x2 + y2
x2 + v2
17.
f(x,y) =
18.
fix y) =
Ax2
19.
fix, y) =
xy
20.
fix, y) =
y
X
21.
fix, y) =
X
y
22.
fix, y) =
3 -
23.
fix, y) =
\x\
In Exercises 24-27, use a computer to provide a portrait of
the given function g(x, y). To do this, (a) use the computer to
help you understand some of the level curves of the function,
and (b) use the computer to graph (a portion of) the surface
Z = g(x, y). In addition, mark on your surface some of the
contour curves corresponding to the level curves you obtained
in part (a). (See Figures 2.10 and 2.11.)
^ 24. g(x, y) = yex
^ 25. g(x,y) = x2 - xy
^ 26. g(x, y) = ix2 + 3y2y-x2-vl
^ sin(2 — x2 — v2)
O 27. g{x, y) = \ .
x £ + yL + 1
28. The ideal gas law is the equation PV = kT, where P
denotes the pressure of the gas, V the volume, T the
temperature, and k is a positive constant.
(a) Describe the temperature T of the gas as a function
of volume and pressure. Sketch some level curves
for this function.
(b) Describe the volume V of the gas as a function
of pressure and temperature. Sketch some level
curves.
29. (a) Graph the surfaces z = x2 and z = y2 ■
(b) Explain how one can understand the graph of the
surfaces z = fix) and z = /(y) by considering
the curve in the ifii-plane given by v = /(«).
(c) Graph the surface in R3 with equation y = x2.
30. Use a computer to graph the family of level curves for
the functions in Exercises 20 and 21 and compare your
results with those obtained by hand sketching. How do
you account for any differences?
31 . Given a function f(x , y), can two different level curves
of / intersect? Why or why not?
In Exercises 32—36, describe the graph of g(x, y, z) by com-
puting some level surfaces. (If you prefer, use a computer to
assist you.)
32. g(x, y,z) = x — 2y + 3z
33. g(x, y,z) = x2 + y2 — z
2.2 | Limits
34. g(x, y,z) = x2 + y2 + z2
35. g(x, y,z) = x2 + 9y2 + 4z2
36. g(x, y, z) = xy - yz
37. (a) Describe the graph of g(x, y, z) = x2 + y2 by
computing some level surfaces.
(b) Suppose g is a function such that the expres-
sion for g(x, y, z) involves only x and y (i.e.,
g(x, y, z) = h(x, yj). What can you say about the
level surfaces of g?
(c) Suppose g is a function such that the expression
for g(x, y, z) involves only x and z. What can you
say about the level surfaces of g?
(d) Suppose g is a function such that the expression
for g(x, y, z) involves only x. What can you say
about the level surfaces of g?
38. This problem concerns the surface determined by the
graph of the equation x2 + xy — xz = 2.
(a) Find a function F(x, y, z) of three variables so
that this surface may be considered to be a level
set of F.
(b) Find a function f(x, y) of two variables so that
this surface may be considered to be the graph of
z = fix, y).
42. x
y2 z2
39. Graph the ellipsoid
x2 y2 2
i.
Is it possible to find a function f(x , y ) so that this ellip-
soid may be considered to be the graph of z = f(x, y)?
Explain.
Sketch or describe the surfaces in R3 determined by the equa-
tions in Exercises 40—46.
2 2
43. x2 + — - — = 0
9 16
40. z
y
41. z2
y
x yi z"
44. — + — = 1
4 16 9
x2 y2 -
45. — + — = z2 - 1
25 16
46. z = y2 + 2
We can look at examples of quadric surfaces with centers
or vertices at points other than the origin by employing a
change of coordinates of the form x = x — xq, y = y — y0,
and z = Z — Zo- This coordinate change simply puts the point
(xo, yo, Zo) of the xyz-coordinate system at the origin of the
xyz-coordinate system by a translation of axes. Then, for ex-
ample, the surface having equation
(x - l)2 (y + 2)2 ,
can be identified by setting x = x — 1, y = y + 2, and z =
Z — 5, so that we obtain
-2 -2
X V _2
which is readily seen to be an ellipsoid centered at (1, —2, 5)
of the xyz-coordinate system. By completing the square in x,
y, or z as necessary, identify and sketch the quadric surfaces
in Exercises 47—52.
47. (* - l)2 + (y + l)2 = (z + 3)2
48. z = 4*2 + (y + 2)2
49. Ax2 + y2 + z2 + 8* = 0
50. Ax2 + y2 - 4z2 + 8x - 4y + 4 = 0
51. *2 + 2y2-6jt-z+10 = 0
52. 9x2 + 4y2 - 36z2 - 8y - 144z = 104
2.2 Limits
As you may recall, limit processes are central to the development of calculus. The
mathematical and philosophical debate in the 1 8th and 1 9th centuries surrounding
the meaning and soundness of techniques of taking limits was intense, questioning
the very foundations of calculus. By the middle of the 19th century, the infamous
"e — 8" definition of limits had been devised chiefly by Karl Weierstrass and
Augustin Cauchy, much to the chagrin of many 20th (and 2 1 st) century students of
calculus. In the ensuing discussion, we study both the intuitive and rigorous mean-
ings of the limit of a function f : X C R" — > Rm and how limits lead to the notion
of a continuous function, our main object of study for the remainder of this text.
98 Chapter 2 I Differentiation in Several Variables
The Notion of a Limit
For a scalar- valued function of a single variable, /:XcR->R, you have seen
the statement
lim f(x) = L
and perhaps have an intuitive understanding of its meaning. In imprecise terms,
the preceding equation (read "The limit of fix) as x approaches a is L.") means
that you can make the numerical value of f(x) arbitrarily close to L by keeping
x sufficiently close (but not equal) to a. This idea generalizes immediately to
functions f:Xc R" — ► Rm . In particular, by writing the equation
lim f(x) = L,
x^a
where f:Xc R" Rm , we mean that we can make the vector f(x) arbitrarily
close to the limit vector L by keeping the vector x g X sufficiently close (but not
equal) to a.
The word "close" means that the distance (in the sense of § 1 .6) between f(x)
and L is small. Thus, we offer a first definition of limit using the notation for
distance.
DEFINITION 2.1 (Intuitive definition of limit) The equation
lim f(x) = L,
x— »a
where HcR"^ Rm, means that we can make ||f(x) — L|| arbitrarily
small (i.e., near zero) by keeping ||x — a|| sufficiently small (but nonzero).
In the case of a scalar- valued function /:XcR" -> R, the vector length
||f(x) — L|| can be replaced by the absolute value |/(x) — L\. Similarly, if / is a
function of just one variable, then ||x — a|| can be replaced by \x — a\.
EXAMPLE 1 Suppose that /: R -> R is given by
1
Figure 2.30 The graph of / of
Example 1.
/(*) =
0
ifx < 1
if x > 1
The graph of / is shown in Figure 2.30. What should limv^i f(x) be? The limit
can't be 0, because no matter how near we make x to 1 (i.e., no matter how small
we take \x — 1 1), the values of x can be both slightly larger and slightly smaller
than 1 . The values of / corresponding to those values of x larger than 1 will
be 2. Thus, for such values of x, we cannot make \f(x) — 0| arbitrarily small,
since, for x > 1, \f(x) — 0| = |2 — 0| = 2. Similarly, the limit can't be 2, since
no matter how small we take \x — l\,x can be slightly smaller than 1. For x < 1,
f(x) = 0 and therefore, we cannot make \f(x) — 2| = |0 — 2| = 2 arbitrarily
small. Indeed it should now be clear that the limit can't be L for any L e R.
Hence, limx^i f(x) does not exist for this function. ♦
EXAMPLE 2 Let f: R2 R2 be defined by f(x) = 5x. (That is, f is five times
the identity function.) Then it should be obvious intuitively that
lim f(x)
x^i+j
lim 5x :
5i + 5j.
2.2 | Limits
Indeed, if we write x = xi + yj, then
||f(x)-(5i + 5j)|| = ||(5xi + 5yj)-(5i + 5j)||
= \\5(x- l)i + 5(y- l)j|| = V25(x- l)2 + 25(j- l)2
= 5j(x - l)2 + (y - 1)2.
This last quantity can be made as small as we wish by keeping
I|X " (i + j)|| = y/{x ~ l)2 + (y - l)2
sufficiently small. ♦
EXAMPLE 3 Now suppose that g:R"^R" is defined by g(x) = 3x. We
claim that, for any a e R",
lim 3x = 3a.
x^a
In other words, we claim that ||3x — 3a || can be made as small as we like by
keeping ||x — a|| sufficiently small. Note that
||3x-3a|| = ||3(x-a)|| =3||x-a||.
This means that if we wish to make ||3x — 3a|| no more than, say, 0.003, then
we may do so by making sure that ||x — a|| is no more than 0.001. If, instead,
we want ||3x — 3a|| to be no more than 0.0003, we can achieve this by keeping
|| x — a|| no more than 0.0001. Indeed, if we want ||3x — 3a|| to be no more than
any specified amount (no matter how small), then we can achieve this by making
sure that ||x — a|| is no more than one-third of that amount.
More generally, if h: R" —> R" is any constant k times the identity function
(i.e., h(x) = kx) and a e R" is any vector, then
lim h(x) = lim kx = ka. ♦
x^a x— >a
The main difficulty with Definition 2.1 lies in the terms "arbitrarily small"
and "sufficiently small." They are simply too vague. We can add some precision
to our intuition as follows: Think of applying the function f: X c R" R"! as
performing some sort of scientific experiment. Letting the variable x take on
a particular value in X amounts to making certain measurements of the input
variables to the experiment, and the resulting value f(x) can be considered to be the
outcome of the experiment. Experiments are designed to test theories, so suppose
that this hypothetical experiment is designed to test the theory that as the input is
closer and closer to a, then the outcome gets closer and closer to L. To verify this
theory, you should establish some acceptable (absolute) experimental error for the
outcome, say, 0.05. That is, you want ||f(x) — L|| < 0.05, if ||x — a|| is sufficiently
small. Then just how small does ||x — a | need to be? Perhaps it turns out that you
must have ||x — a|| < 0.02, and that if you do take ||x — a|| < 0.02, then indeed
||f(x) — L|| < 0.05. Does this mean that your theory is correct? Not yet. Now,
suppose that you decide to be more exacting and will only accept an experimental
error of 0.005 instead of 0.05. In other words, you desire ||f(x) — L|| < 0.005.
Perhaps you find that if you take ||x — a|| < 0.001, then this new goal can be
achieved. Is your theory correct? Well, there's nothing sacred about the number
0.005, so perhaps you should insist that ||f(x) — L|| < 0.001, or that ||f(x) — L|| <
0.00001. The point is that if your theory really is correct, then no matter what
(absolute) experimental error e you choose for your outcome, you should be
able to find a "tolerance level" 8 for your input x so that if ||x — a|| < <5, then
Chapter 2 | Differentiation in Several Variables
|| f(x) — L || < e. It is this heuristic approach that motivates the technical definition
of the limit.
DEFINITION 2.2 (RIGOROUS DEFINITION OF limit) Let f : X C R" Rm
be a function. Then to say
lim f(x) = L
x— *a
means that given any e > 0, you can find a S > 0 (which will, in general,
depend on e) such that if x e X and 0 < ||x — a|| < <5, then ||f(x) — L|| < e.
The condition 0 < ||x — a|| simply means that we care only about values
f(x) when x is near a, but not equal to a. Definition 2.2 is not easy to use in
practice (and we will not use it frequently). Moreover, it is of little value insofar
as actually evaluating limits of functions is concerned. (The evaluation of the
limit of a function of more than one variable is, in general, a difficult task.)
EXAMPLE 4 So that you have some feeling for working with Definition 2.2,
let's see rigorously that
lim (3x - 5y + 2z) = 12
tw)->- (1,-1,2)
(as should be "obvious"). This means that given any number e > 0, we can find
a corresponding S > 0 such that
ifO < \\(x,y,z)-(l, -1,2)|| < 8, then \3x -5y + 2z- 12| < e.
(Note the uses of vector lengths and absolute values.) We'll present a formal
proof in the next paragraph, but for now we'll do the necessary background
calculations in order to provide such a proof. First, we need to rewrite the two
inequalities in such a way as to make it more plausible that the e-inequality could
arise algebraically from the S -inequality. From the definition of vector length, the
(5-inequality becomes
0 < y/(x- l)2 + (y + l)2 + (z-2)2 < 8.
If this is true, then we certainly have the three inequalities
V(jc - l)2 = \x - 1| <S,
y(V+l)2 = |y+l| <S,
y/(z - 2)2 = \z-2\ < S.
Now, rewrite the left side of the e-inequality and use the triangle inequality (2)
of §1.6:
\3x -5y + 2z- 12| = |3(* - 1) - 5(y + 1) + 2(z - 2)|
< |3(x - 1)| + \5(y + 1)| + \2(z - 2)|
= 3\x - 1| +5|y + 1| +2|z-2|.
Thus, if
0< \\(x,y,z)- (1,-1, 2)|| <S,
then
\x- 1| <8, \y+l\ <S, and \z-2\ < 8,
so that
\3x-5y + 2z- 12| < 3\x - 1| + 5\y + 1| + 2\z - 2\
< 38 + 58 + 28 = 108.
2.2 | Limits 101
If we think of S as a positive quantity that we can make as small as desired, then
105 can also be made small. In fact, it is 10<5 that plays the role of e.
Now for a formal, "textbook" proof: Given any e > 0, choose 8 > 0 so that
S < e/10. Then, if
0< \\(x,y,z)- (1,-1, 2)|| <S,
it follows that
|x-l|<<5, |y+l|<<5, and \z-2\<8,
so that
\3x-5y + 2z- 12| < 3|x - 1| + 5\y + 1|+ 2|z - 2|
< 3<5 + 55 + 25
= 105 < 10— = e.
~ 10
Thus, lim(x,VjZ)^(1,_ii2)(3x — 5y + 2z) = 12, as desired. ♦
Using the same methods as in Example 4, you can show that
lim(fliJfi + (22*2 + ■ ■ ■ + anxn) = a\b\ + 02^2 + ■ ■ ■ + a„bn
for any cij, i = 1,2, n.
Some Topological Terminology
Before discussing the geometric meaning of the limit of a function, we need to in-
troduce some standard terminology regarding sets of points in R" . The underlying
geometry of point sets of a space is known as the topology of that space.
Recall from §2.1 that the vector equation ||x — a|| = r, where x and a are
in R3 and r > 0, defines a sphere of radius r centered at a. If we modify this
equation so that it becomes the inequality
Hx-a||<r, (1)
then the points x e R3 that satisfy it fill out what is called a closed ball shown in
Figure 2.31. Similarly, the strict inequality
||x-a||<r (2)
describes points x e R3 that are a distance of less than r from a. Such points
determine an open ball of radius r centered at a, that is, a solid ball without the
boundary sphere.
There is nothing about the inequalities (1) and (2) that tie them to R3. In fact,
if we take x and a to be points of R", then (1) and (2) define, respectively, closed
and open n -dimensional balls of radius r centered at a. While we cannot draw
sketches when n > 3, we can see what (1) and (2) mean when n is 1 or 2. (See
Figures 2.32 and 2.33.)
y y
X
Figure 2.32 The closed and open balls (disks) in R2 defined by [|x — a[| < r and
Chapter 2 | Differentiation in Several Variables
X
Figure 2.34 The graph of X.
-s^ —
Figure 2.37 The set X of
Example 7.
Figure 2.33 The closed and open balls (intervals) in R
defined by \x — a\ < r and \x — a\ < r.
DEFINITION 2.3 A set X C R" is said to be open in R" if, for each point
x e X, there is some open ball centered at x that lies entirely within X. A
point x e R" is said to be in the boundary of a set X c R" if every open ball
centered at x, no matter how small, contains some points that are in X and
also some points that are not in X. A set X cR" is said to be closed in R"
if it contains all of its boundary points. Finally, a neighborhood of a point
x g X is an open set containing x and contained in X.
It is an easy consequence of Definition 2.3 that a set X is closed in R" precisely
if its complement R" — X is open.
EXAMPLE 5 The rectangular region
X = {(x, y) g R2 | -1 < x < 1, -1 < y < 2}
is open in R2. (See Figure 2.34.) Each point in X has an open disk around it
contained entirely in the rectangle. The boundary of X consists of the four sides
of the rectangle. (See Figure 2.35.) ♦
X
Figure 2.35 Every open disk
about a point on a side of rectangle
X of Example 5 contains points in
both X and R2 — X.
Figure 2.36 The set X
of Example 6 consists of
the nonnegative
coordinate axes.
EXAMPLE 6 The set X consisting of the nonnegative coordinate axes in R3 in
Figure 2.36 is closed since the boundary of X is just X itself. ♦
EXAMPLE 7 Don't be fooled into thinking that sets are always either open or
closed. (That is, a set is not a door.) The set
X = {(x, y) g R2 | 0 < x < 1, 0 < y < 1}
shown in Figure 2.37 is neither open nor closed. It's not open since, for example,
the point Q, 0) that lies along the bottom edge of X has no open disk around it
that lies completely in X. Furthermore, X is not closed since the boundary of X
includes points of the form (x, 1) for 0 < x < 1 (why?), which are not part of X.
2.2 | Limits 103
The Geometric Interpretation of a Limit
Suppose that He R" — » Rm. Then the geometric meaning of the statement
lim f(x) = L
\ Ml
is as follows: Given any e > 0, you can find a corresponding S > 0 such that if
points x e X are inside an open ball of radius 8 centered at a, then the correspond-
ing points f(x) will remain inside an open ball of radius e centered at L. (See
Figure 2.38.)
y
Figure 2.38 Definition of a limit: Given an open ball 5e centered at L (right), you
can always find a corresponding ball Bg centered at a (left), so that points in Bs n X
are mapped by f to points in Be .
We remark that for this definition to make sense, the point a must be such
that every neighborhood of it in R" contains points xeX distinct from a. Such
a point a is called an accumulation point of X. (Technically, this assumption
should also be made in Definition 2.2.) A point a e X is called an isolated point
of X if it is not an accumulation point, that is, if there is some neighborhood of a
in R" containing no points of X other than a.
From these considerations, we see that the statement limx^a f(x) = L really
does mean that as x moves toward a, f(x) moves toward L. The significance of
the "open ball" geometry is that entirely arbitrary motion is allowed.
EXAMPLE 8 Let /: R2 - {(0, 0)} -> R be defined by
/(*. y) =
y
x2 + y2
Let's see what happens to / as x = (x, y) approaches 0 = (0, 0). (Note that / is
undefined at the origin, although this is of no consequence insofar as evaluating
limits is concerned.) Along the x-axis (i.e., the line y = 0), we calculate the value
of / to be
,•2
f(x, 0) =
0
0
1.
Thus, as x approaches 0 along the line y = 0, the values of / remain constant,
and so
lim /(x)=l.
x— > 0 along y=0
Along the j-axis, however, the value of / is
m y)
0-y2
0 + y2
1 04 Chapter 2 | Differentiation in Several Variables
Hence,
lim /(x) = -l.
>0 along x=0
Indeed, the value of / is constant along each line through the origin. Along the
line y = mx, m constant, we have
f(x, mx)
Therefore,
x2 — m2x2 x2(l — m2) 1 — m2
x2+m2x2 x2(\ + m2) 1 + m2
1 - m2
lim f(x)
x^O along y=mx 1 -|- m
As a result, the limit of / as x approaches 0 does not exist, since / has different
"limiting values" depending on which direction we approach the origin. (See
Figure 2.39.) That is, no matter how close we come to the origin, we can find
points x such that f(x) is not near any number LeR, (In other words, every
open disk centered at (0, 0), no matter how small, is mapped onto the interval
[ — 1 , 1].) If we graph the surface having equation
2 2
x — y
x2 + y2
(Figure 2.40), we can see quite clearly that there is no limiting value as x ap-
proaches the origin. ♦
y
R2
Figure 2.39 The function f{x, y) = (x2 - y2)/(x2 + y2) of Figure 2.40 The graph of f(x, y) =
Example 8 has value 1 along the jc-axis and value - 1 along the (* - y )/(* + y1) of Example 8.
y-axis (except at the origin).
Warning Example 8 might lead you to think you can establish that
lim^a f(x) = L by showing that the values of f as x approaches a along straight-
line paths all tend toward the same value L. Although this is certainly good
evidence that the limit should be L, it is by no means conclusive. See Exercise 23
for an example that shows what can happen.
EXAMPLE 9 Another way we might work with the function f(x, y) = (x2 —
y2)/(x2 + y2) of Example 8 is to rewrite it in terms of polar coordinates. Thus,
let x = r cos 9, y = r sin 9. Using the Pythagorean identity and the double angle
2.2 | Limits 105
formula for cosine, we obtain, for r ^ 0, that
x2 - y2 r2 cos2 9 - r2 sin2 9 r2(cos2 9 - sin2 9) cos 29
= ~ = — cos 2$
x2 + y2 r2 cos2 (9 + r2 sin2 (9 r2(cos2 6» + sin2 6») 1
That is, for r / 0,
f(x,y) = f(r cos0, r sin#) = cos 20.
Moreover, to evaluate the limit of / as (x, y) approaches (0, 0), we only must
have r approach 0; there need be no restriction on 9. Therefore, we have
lim f(x, y) = limcos2# = cos2#.
This result clearly depends on 9. For example, if 9 = 0 (which defines the x-axis),
then
lim cos 29 = 1,
r^O along 6 = 0
while if 9 = tt/4 (which defines the line y = x), then
lim cos 29 = 0.
r^0 along 6 = jr/4
Thus, as in Example 8, we see that lim(A V)^(o,o) f(x, y) fails to exist. ♦
EXAMPLE 10 We use polar coordinates to investigate lim(A>Y)^(o,o) f(x, y),
where f(x , y) = (x3 + x5)/(x2 + y2).
We first rewrite the expression (x3 + x5)/(x2 + y2) using polar coordinates:
x3 + x5 r3 cos3 0 + r5 cos5 1
x2 + y2 r2 cos2 6> + r2 sin
Now — 1 < cos# < 1, which implies that
r(cos3 9 + r2 cos5 9).
Hence,
■ 1 - r2 < cos3 9 + r2 cos5 9 < 1 + r2.
-r(\ + r2)< f(x,y)<r(l+r2).
As r -> 0, both the expressions — r(l + r2) and r(l + r2) approach zero. Hence,
we conclude that lim(x,v)^(o,o) /(x, y) = 0, since / is squeezed between two
expressions with the same limit. ♦
Properties of Limits
One of the biggest drawbacks to Definition 2.2 is that it is not at all useful for
determining the value of a limit. You must already have a "candidate limit" in
mind and must also be prepared to confront some delicate work with inequalities
to use Definition 2.2. The results that follow (which are proved in the addendum
Chapter 2 | Differentiation in Several Variables
to this section), plus a little faith, can be quite helpful for establishing limits, as
the subsequent examples demonstrate.
THEOREM 2.4 (Uniqueness OF limits) If a limit exists, it is unique. That is,
let f: X C R" Rm. If limx^a f(x) = L and limx^a f(x) = M, then L = M.
THEOREM 2.5 (Algebraic properties) Let F,G:Xc R" -> R" be vector-
valued functions, f,g:X^ R" -> R be scalar-valued functions, and let k € R
be a scalar.
1. If limx^a F(x) = L and limx^a G(x) = M,
then limx^a(F + G)(x) = L + M.
2. If linwa F(x) = L, then limx^a k¥(x) = kh.
3. If limx^a f(x) = L and limx^a g(x) = M, then limx^a(/g)(x) = LM.
4. If limx^a /(x) = L, g(x) / 0 for x e X, and limx^a g(x) = M/ 0, then
linwa(//g)(x) = L/M.
There is nothing surprising about these theorems — they are exactly the same
as the corresponding results for scalar- valued functions of a single variable. More-
over, Theorem 2.5 renders the evaluation of many limits relatively straightforward.
EXAMPLE 1 1 Either from rigorous considerations or blind faith, you should
find it plausible that
lim x = a and lim v = b.
(x,y)^(a,b) (x,y)^(a,b)
From these facts, it follows from Theorem 2.5 parts 1, 2, and 3 that
lim (x2 + 2xy - y3) = a2 + lab - b3 ,
(x,y)-*(a,b)
because, by part 1 of Theorem 2.5,
lim (x2 + 2xy - y3) = limx2 + lim2jcy + lim(-y3)
(x,y)-^(a,b)
and, by parts 2 and 3,
lim (x2 + Ixy - y3) = (limx)2 + 2(limx)(limy) - (limy)3
(x,y)-*(a,b)
so that, from the facts just cited,
lim (x2 + 2xy - y3) = a2 + lab - b3 . ♦
(x,yy+(a,b)
EXAMPLE 12 More generally, a polynomial in two variables x and y is any
expression of the form
d d
p(x,y) = J2J2ckixkyl,
k=0 1=0
where d is some nonnegative integer and c« 6 R for k, I = 0, . . . , d. That is,
p(x, y) is an expression consisting of a (finite) sum of terms that are real number
coefficients times powers of x and y. For instance, the expression x2 + 2xy — y3
in Example 11 is a polynomial. For any (a, b) e R2, we have, by part 1 of
2.2 | Limits 107
Theorem 2.5,
d d
lim p(x, v) = y^y^ lim (ctixkyl)
so that, from part 2,
d d
lim p(x, y) = 7^ 7^ cm lim xkyl
(x,y)->(a,b) f^0 (x,y)^(a,b)
and, from part 3,
d d
lim p(x, j) = VV cH(limx*)(limyz)
= ±±cklakbK
k=0 1=0
Similarly, a polynomial in n variables x\, x%, . . . , xn is an expression of the form
p(x\,x2, . . . ,xn) = ^ c*:i..4nxf1x22 • • •**"
fei,...,ft„=0
where is some nonnegative integer and c*,...^ 6 R for fci, . . . , fc„ = 0, . . . , d.
For example, a polynomial in four variables might look like this:
p{X\, . . . , Xa) = 1x\x-i + — 7X3X4.
Theorem 2.5 implies readily that
lim -k, X\ x2 ■■■ xn" = Ckv-Ka\ a2 ' ' ' a„ ■
EXAMPLE 13 We evaluate lim * +Xy + ] .
(x,yy+(-l,0) x2y - 5xy + y2 + 1
Using Example 12, we see that
lim x2 + xv + 3 = 4,
(x,yH-(-l,0)
and
lim x2y - 5xy + y2 + 1 = 1(/ 0).
(A-,y)^<-1.0) '
Thus, from part 4 of Theorem 2.5, we conclude that
lim *2 + '? + 3 =1 = 4.
(x,y)^(-i,0) x2y - 5xy + y2 + 1 1
EXAMPLE 14 Of course, not all limits of quotient expressions are as simple
to evaluate as that of Example 13. For instance, we cannot use Theorem 2.5 to
evaluate
2 4
x — y
lim — j (3)
since ]jm(Xtyy+(o,o)(x2 + y4) = 0. Indeed, since lim^^-^o.o^x2 — y4) = 0 as
well, the expression (x2 — y4)/(x2 + y4) becomes indeterminate as (x, y) ->
(0, 0). To see what happens to the expression, we note that
x2 4 x2
lim — = lim — = 1,
.1-^0 along y=0 X2 + y4 x-*0 X2
Chapter 2 | Differentiation in Several Variables
while
lim
>0 along
=o x2 + y4
= lim
v4
Thus, the limit in (3) does not exist. (Compare this with Example 8.) ♦
The following result shows that evaluating the limit of a function
f : X c R" —> R" is equivalent to evaluating the limits of its (scalar- valued) com-
ponent functions. First recall from §2.1 that f(x) may be rewritten as (/i(x),
/2(x), . . . , /m(x)).
THEOREM 2.6 Suppose He R" R"1 is a vector-valued function. Then
limx^a f(x) = L, where L = (Li, . . . , Lm), if and only if limx^a ft(x) = Li for
i = 1, .... m.
EXAMPLE 15 Consider the linear mapping f: R" -■>• R" defined by f(x) = Ax,
where A = (a,;) is an m x n matrix of real numbers. (See Example 5 of §1.6.)
Theorem 2.6 shows us that
for any h = (b\
lim f(x) = Ab
x— >b
, bn) in R" . If we write out the matrix multiplication, we have
f(x) = Ax =
flu
«21
_ Ami
flln
fl2;i
-Vl
A'2
mn _1 l_ -^n _
aUX\ + (312^2 +
Cl2\X\ + a22*2 +
~f" Cl\n%n
flml-^-1 ~T~ @m2X2 1***1 0.mnXn
Therefore, the tth component function of f is
= anxi + a,-2x2 H h fli«x„.
From Example 4, we have that
lim yi(x) = fln^i + fl/2^2 H h a,>A
x^b
for each /. Hence, Theorem 2.6 tells us that the limits of the component functions
fit together to form a limit vector. We can, therefore, conclude that
limf(x) = (lim/1(x), .
x^b x^b
= (aubi H
anb] +
a2\b\ +
, lim /m(x))
\ > b
■ ci\nbn, . . . , am\b\ +
+ a\nbn
+ a2nbn
®mnb\\)
am\b\ + • • ■ + a„,„b„
once we take advantage of matrix notation.
Ab,
2.2 | Limits 109
Figure 2.41 The graph of a
continuous function.
Figure 2.44 The graph of a
continuous function f(x, y).
Continuous Functions
For scalar-valued functions of a single variable, one often adopts the following
attitude toward the notion of continuity: A function /: X C R is continuous
if its graph can be drawn without taking the pen off the paper. By this criterion,
Figure 2.41 describes a continuous function y = f(x), while Figure 2.42 does not.
Figure 2.42 The graph of a
function that is not continuous.
Figure 2.43 The graph of /
where f(x, y) = 0 if both x > 0
and y > 0, and where f(x, y) = 1
otherwise.
We can try to extend this idea to scalar-valued functions of two variables: A
function /: X C R2 — R is continuous if its graph (in R3) has no breaks in it.
Then the function shown in Figure 2.43 fails to be continuous, but Figure 2.44
depicts a continuous function. Although this graphical approach to continuity is
pleasantly geometric and intuitive, it does have real and fatal flaws. For one thing,
we can't visualize graphs of functions of more than two variables, so how will we
be able to tell in general if a function /:XcR"^ Rm is continuous? Moreover,
it is not always so easy to produce a graph of a function of two variables that is
sufficient to make a visual determination of continuity. This said, we now give a
rigorous definition of continuity of functions of several variables.
DEFINITION 2.7 Let f: X C R" -> R"! and let a e X. Then f is said to be
continuous at a if either a is an isolated point of X or if
limf(x) = f(a).
x->a
If f is continuous at all points of its domain X, then we simply say that f is
continuous.
EXAMPLE 16 Consider the function /: R2
x2 + xy — 2y2
f(x,y)= '
x2 + y2
0
-> R defined by
if (x, y) # (0, 0)
if (x, y) = (0, 0)
Therefore, f(0, 0) = O^utlim^^^o) f(x, y) does not exist. (To see this, check
what happens as (x, y) approaches (0,0) first along y = 0 and then along x = 0.)
Hence, / is not continuous at (0,0). ♦
It is worth noting that Definition 2.7 is nothing more than the "vectorized"
version of the usual definition of continuity of a (scalar-valued) function of one
variable. This definition thus provides another example of the power of our vector
notation: Continuity looks the same no matter what the context.
Chapter 2 | Differentiation in Several Variables
One way of thinking about continuous functions is that they are the ones whose
limits are easy to evaluate: When f is continuous, the limit of f as x approaches
a is just the value of f at a. It's all too tempting to get into the habit of behaving
as if all functions are continuous, especially since the functions that will be of
primary interest to us will be continuous. Try to avoid such an impulse.
EXAMPLE 1 7 Polynomial functions in n variables are continuous. Example 12
gives a sketch of the fact that
lim y* ckv..kix\1 ■ ■ ■ xl" = y~] ch...kaa^ ■ ■ ■ a*",
where x = (xi, . . . , xn) and a = (#i, . . . , an) are in R". If /: R" —> R is defined
by
/(x) = X>*,...fc*?'---#.
then the preceding limit statement says precisely that / is continuous at a. ♦
EXAMPLE 18 Linear mappings are continuous. If f: R" — > R'" is denned by
f(x) = Ax, where A is an m x n matrix, then Example 1 5 establishes that
lim f(x) = Ab = f(b)
for all b e R". Thus, f is continuous. ♦
The geometric interpretation of the e — S definition of a limit gives rise to a
similar interpretation of continuity at a point: f: X C R" —> R"! is continuous at
a point a £ X if, for every open ball B€ in R" of radius e centered at f(a), there
is a corresponding open ball Bs in R" of radius 8 centered at a such that points
x e X inside B& are mapped by f to points inside B(. (See Figure 2.45.) Roughly
speaking, continuity of f means that "close" points in X c R" are mapped to
"close" points inRffl.
In practice, we usually establish continuity of a function through the use
of Theorems 2.5 and 2.6. These theorems, when interpreted in the context of
2.2 | Limits 1 1 1
continuity, tell us the following:
• The sum F + G of two functions F, G: X C R" — > Rm that are continuous
at a e X is continuous at a.
• For all & € R, the scalar multiple kF of a function F:XcR"->Rffl that
is continuous at a £ X is continuous at a.
• The product fg and the quotient f/g(gj^ 0) of two scalar- valued functions
f,g:XC. R" R that are continuous ata e I are continuous at a.
• F:XcR"-> R'" is continuous at a e X if and only if its component
functions F* : X C R" — »• R, j = 1 , . . . , m are all continuous at a.
EXAMPLE 1 9 The function f: R2 -> R3 denned by
f(x, y) = (x + y, x2y, y sin(xy))
is continuous. In view of the remarks above, we can see this by checking that the
three component functions
fi(x,y) = x + y, f2(x,y) = x2y, and f3(x, y) = y sin(xy)
are each continuous (as scalar-valued functions). Now f\ and fi are continuous,
since they are polynomials in the two variables x and y. (See Example 17.) The
function is the product of two further functions; that is,
Mx, y) = g(x, y)h(x, y),
where g{x, y) = y and h(x, y) = sin(xy). The function g is clearly continuous.
(It's a polynomial in two variables — one variable doesn't appear explicitly!) The
function h is a composite of the sine function (which is continuous as a function
of one variable) and the continuous function p(x, y) = xy. From these remarks,
it's not difficult to see that
lim h(x, y) = lim sin(p(x, y))
(x,y)-+(a,b) (x,y)^(a,b)
= sin ( lim p(x, y) I ,
since the sine function is continuous. Thus,
lim h(x, y) = sin p(a, b) = h(a, b),
because p is continuous. Thus, h, hence fa, and, consequently, fare all continuous
on all of R2. ♦
The discussion in Example 19 leads us to the following general result, whose
proof we omit:
THEOREM 2.8 Iff: XcR"^ R"< andg: Y C R" RP are continuous func-
tions such that range fey, then the composite function g o f: X C R" — > Rp is
defined and is also continuous.
Chapter 2 | Differentiation in Several Variables
Addendum: Proofs of Theorems 2.4, 2.5, 2.6, and 2.8
For the interested reader, we establish the various results regarding limits of
functions that we used earlier in this section.
Proof of Theorem 2.4 The statement limx^a f(x) = L means that, given any e >
0, we can find some Si > 0 such that if x 6 X and 0 < ||x — a|| < Si, then ||f(x) —
L|| < e/2. (The reason for writing e/2 rather than e will become clear in a
moment.) Similarly, limx^a f(x) = M means that, given any e > 0, we can find
some S2 > 0 such that if x e X and 0 < ||x - a|| < S2, then ||f(x) - M|| < e/2.
Now let S = min^!, £2); that is, we set S to be the smaller of Si and S2. If
x e X and 0 < ||x - a|| < 8, then both ||f(x) - L|| and ||f(x) - M|| are less than
e /2 so that, using the triangle inequality, we have
||L-M|| = ||(L-f(x)) + (f(x)-M)||
< ||L-f(x)|| + ||f(x)-M|| <| + | = 6.
This shows that the quantity ||L — M|| can be made arbitrarily small; thus, it
follows that L - M = 0. Hence, L = M. ■
Proof of Theorem 2.5 To establish part 1, note that if limx^aF(x) = L, then
given any e > 0, we can find a Si > 0 such that if x e X and 0 < ||x — a|| <
Si, then || F(x) — L|| < e/2. Similarly, if limx^a G(x) = M, then we can find a
S2 > 0 such that if x 6 X and 0 < ||x - a|| < S2, then ||G(x) - M|| < e/2. Now
let i5 = min((5i, ^2). Then if x e X and 0 < ||x — a|| < <5, the triangle inequality
implies that
||(F(x) + G(x)) - (L + M)|| < ||F(x) - L|| + ||G(x) - M\\ <"-+"-= e.
Hence, limx^a (F(x) + G(x)) = L + M.
To prove part 2, suppose that e > 0 is given. If limx^a F(x) = L, then we can
find a <5 > 0 such that if x £ X and 0 < ||x - a|| < S, then ||F(x) - L|| < e/\k\.
Therefore,
\\k¥(x) - *L|| = |*| ||F(x) - L|| < |^| -1 = e,
\k\
which means that limx^a k¥(x) = kL,. (Note: If k = 0, then part 2 holds trivially.)
To establish the rule for the limit of a product of scalar- valued functions (part
3), we will use the following algebraic identity:
/(x)g(x) -LM = (/(x) - L)(g(x) — M) + L(g(x) — M) + M(f(x) - L).
(4)
If limx^a f(x) = L, then, given any e > 0, we can find Si > 0 such that if x e X
and 0 < ||x — a|| < Si, then
|/(x)-L| < Vi\
Similarly, if limx^a g(x) = M, we can find S2 > 0 such that if x e X and 0 <
||x — a|| < S2, then
\g(x)-M\ < Je.
Let 8 = min((5i , ^2). If x e X and 0 < ||x — a|| < 8, then
|(/(x) - L)(g(x) -M)\<sTe-^e = e.
2.2 | Limits 1 1 3
This means that limx^a(/(x) — L)(g(x) — M) = 0. Therefore, using (4) and parts
1 and 2, we see that
lim(/(x)g(x) - LM) = lim(/(x) - L)(g(x) -M) + L lim(g(x) - M)
x^- a x-> a x-^- a
+ M lim(/(x) - L)
x— >a
= 0 + 0 + 0 = 0.
Since limx^a f(x)g(x) = limx^a((/(x)^(x) — LM) + LM), the desired result
follows from part 1 .
The crux of the proof of part 4 is to show that
lim
1
1
~M
x^a g(x)
Once we show this, the desired result follows directly from part 3 :
/(x)
lim
x^a g(X)
Note that
= lim /(X)
1
M
1
S(x)
S(x)
_ \M - g(x)\
\Mg(x)\
1 _ L
~ ~M ~ ~M
and, by the triangle inequality, that
\M\ = \M- g(x) + g(x)| <\M- g(x)\ + \g(x)
(5)
If limx^.a g(x) = M, then, given any e > 0, we can find 8\ such that if x e X and
0 < ||x- a|| < <5!,then
M2
\g(x)-M\ <—e.
We can also fmd 82 such that if x e X and 0 < ||x — a|| < 82, then |g(x) — M\ <
\M\/2 and, hence, using (5), that
|M|
\M\ < + 1*001 \g(x)\ >
\M\
1
Now let 8 = min(<5i , ^2). If x e X and 0
1
g(x)
1
M
|M-g(x)|
|Mg(x)|
1 2 M2
2 l#(x)|
x — a|| < 8, then
1 |M-g(x)|
" IM|
<
|M|
lg(x)l
< |M| |M| 2 e €'
Proof of Theorem 2.6 Note first that, for i = 1, . . . , m,
\fi(x) ~L,\< V(/1(x)-L1)2 + --- + (/m(x)-Lm)2 = ||f(x) - L||
(6)
If limx^a f(x) = L, then given any e > 0, we can find a 5 > 0 such that if x e
X and 0 < ||x - a|| < 8, then ||f(x) - L|| < e. Hence, (6) implies that |/;(x) -
Lf\ < € for i = 1, . . . , m, which means that limx^a /,(x) = L;.
Conversely, suppose that limx^a /i(x) = L,- for i = 1, . . . , m. This means
that, given any e > 0, we can find, for each i , a 8t > 0 such that if x e X and
14 Chapter 2 | Differentiation in Several Variables
0 < ||x — a|| < 8i, then |/}(x) — L,| < e/^Jm. Set 8 = min(<5i, . . . , 8m). Then if
x e X and 0 < ||x — a|| < 8, we see that (6) implies
||f(x)-L|| < J- + = m- = €.
V m m V m
Thus, limx^a f(x) = L. ■
Proof of Theorem 2.8 We must show that the composite function g o f is con-
tinuous at every point a e X. If a is an isolated point of X, there is nothing to
show. Otherwise, we must show that limx^a(g o f)(x) = (go f)(a).
Given any e > 0, continuity of g at f(a) implies that we can find some y > 0
such that if y e range f and 0 < ||y — f(a)|| < y then
||g(y)-g(f(a))|| <6.
Since f is continuous at a, we can find some <5 > 0 such that if x e X and 0 <
|| x — a|| < 8, then
||f(x)-f(a)|| < y.
Therefore, if x e X and 0 < ||x — a|| < 8, then
||g(f(x))-g(f(a))|| <e. m
2.2 Exercises
In Exercises 1—6, determine whether the given set is open or
closed (or neither).
1. {(x,y)eR2 | 1 < x2 + y2 < 4}
2. {(x,y)eR2 | 1 <x2+y2 < 4}
3. {(x,y)eR2 | 1 < x2 + y2 < 4}
4. {(x, y, z) e R3 | 1 < x2 + y2 + z2 < 4}
5. {(x, y) e R2 | -1 < x < 1} U {(x, y) e R2 |
6. {(x,y,z)eR3 | 1 < x2 + y2 < 4}
2}
Evaluate the limits in Exercises 7—21, or explain why the limit
fails to exist.
x2 + 2xy + yz + z3 + 2
7. lim
(*,.y,z)-H0,0,0)
8. lim
l.v|
(x,y)->(0,0) jxl _|_ yl
9. lim — -
(x.y)^r(P.O) x2 + y2
10. lim
(jt.yH-ro.O) x + y + 2
2x2 + y2
1 1 . lim
(jt,y)-+(0.0) X2 + y2
2x2 + y2
12. lim
(x,y)^(-l,2) X2 + y2
x2 + 2xy + y2
u-O -to ui x + y
xy
13. lim
14. lim
(*,;>>)-K0,0) x2 + y2
x4-y4
15. lim — ■—
(jr,y)-<-(0,0) x1 + y1
X2
16. lim — -
(x,y)^(0,0) X2 + y2
17.
lim
x — xy
(jc,y)->(o.o)>*54> -Jx -
x2-y2-4x+4
1 8. lim — —
(x,y)^(2,0) x2 +y2 -4x +4
19.
20.
21.
lim exz cos y — x
(x,y,z)^(0,^fit.l)
2x2 + 3y2 + z2
(jc,y,t)->-(0,0,a) X2 + y2 + Z2
xy — xz + yz
lim
lim
(jc,y,t)-K0,0,a) x2 + y2 + z2
sin©
22. (a) What is lim ?
(b) What is lim Sm(x + y)?
(x,y)^(0,0) x + y
(c) What is lim
(x,y)^(0,0) xy
2.2 | Exercises 115
23. Examine the behavior of fix, y) = x4y4/(x2 + y4)3
as (x, y) approaches (0, 0) along various straight lines.
From your observations, what might you conjecture
lim(XJ,)_i.(oio) f(x, y) to be? Next, consider what hap-
pens when (x, y) approaches (0, 0) along the curve
x = y2. Does lim^- yj^^o) fix, y) exist? Why or why
not?'
In Exercises 24—27, (a) use a computer to graph z = f(x, y);
(b) use your graph in part (a) to give a geometric discussion
as to whether lim^ ^^^o) f(x, y) exists; (c) give an analytic
(i.e., nongraphical) argument for your answer in part (b).
4x2 + 2xy + 5 y2
^ 24. f(x, y) :
^25. f(x,y)-.
^ 26. f{x, y) :
27. f(x, y) :
3x2 + 5y2
y
x2 + y2
„5
xy
x2 + v1
x sin — if y ^ 0
y
o
ify = 0
Some limits become easier to identify if we switch to a different
coordinate system. In Exercises 28-33 switch from Cartesian to
polar coordinates to evaluate the given limits. In Exercises 34—
37 switch to spherical coordinates.
28.
29.
30.
31.
32.
33.
34.
35.
36.
37.
lim
lim
*,v>^(0.0)
lim
lim — -
z,}>)->-(0,0) x2 + y2
x2
lim — r
x,y)-y(0,0) X2 + y2
x2 + xy + y2
x2+y2
x5 + y4 - 3x3y + 2x2 + 2y2
x2 + y2
x + y
lim
lim
2
x y
x,j,z)^(0,o,o) x2 + y2 + z2
xyz
lim —
x,y,z)^(0,0,0) x2 + y2 + z2
lim
x2 + y2
*,y,z)-K0,0.0) Jx2 + y2 +z2
XZ
lim — = -
>,y,z)^-(0,0,0) X2 + V2 + Z2
In Exercises 38—45, determine whether the functions are con-
tinuous throughout their domains:
38. fix,y) = x2 + 2xy - y1
39. fix, y, z) = x2 + 3xyz + yz3 +2
40. g(x,y):
x2 + \
41. h(x, y) = cos
x2-y2
x2 + \
42. fix, y) = cos2 x — 2 sin2 xy
if(x,y)/(0,0)
0 if(x,y) = (0,0)
x3 + x2 + xy2 + y
43. fix, y) :
x2-y2
x^Ty2
44. g(x,y) .
x2 + y2
2
45. F(x,y,z)= [xz + 3xy
if(x,y)/(0, 0)
if(x,y) = (0, 0)
xy
2x2 + y4 + 3 ' \v2 + l
46. Determine the value of the constant c so that
' x3 +xy2 +2x2 +2y2
g(x, y) =
is continuous.
x + y2
c
if ix, y) # (0, 0)
if ix, y) = (0, 0)
47. Show that the function /: R3 — > R given by f(\) =
(2i — 3j + k) • x is continuous.
48. Show that the function f: R3 — > R3 given by f(x) =
(6i — 5k) x x is continuous.
Exercises 49—53 involve Definition 2.2 of the limit.
49. Consider the function fix) = 2x — 3.
(a) Show that if \x - 5| < 5, then |/(jc) - 7| < 25.
(b) Use part (a) to prove that limA-^5 fix) = 7.
50. Consider the function fix, y) = 2x — lOy + 3.
(a) Show that if ||(jc, y) - (5, 1)|| < S, then \x - 5| <
S and |y — 1 1 < S.
(b) Use part (a) to show that if ||(jc, y) - (5, 1)|| < S,
then|/(A,y)- 3| < 125.
(c) Show that lim(x },)^(5i) fix, y) = 3.
51 . If A, B, and C are constants and fix, y) = Ax + By +
C, show that
lim fix, y) = fix0, y0) = Ax0 + By0 + C.
(jc,y)-s-(x0,>'o)
Chapter 2 | Differentiation in Several Variables
52. In this problem, you will establish rigorously that
lim — = 0.
(x,y)M0,0) x2 + y2
(a) Show that \x\ < \\{x,y)\\ and \y\ < \\(x,y)\\.
(b) Showthat |jc3 + y3\ < 2(x2 + y2fl2. (Hint: Begin
with the triangle inequality, and then use part (a).)
(c) Show that if 0 < ||(jc, y)|| < S, then \(x3 + y3)/
(x2 + y2)\ < 28.
(d) Now prove that \im(x v)^(0 o)(-*3 + v3)/(x2 +
y1) = o.
53. (a) If a and b are any real numbers, show that 2\ab\ <
a2 + b2.
(b) Let
(x2- y2\
Use part (a) to show that if 0 < ||(jc, y)\\ < S, then
\f(x,y)\ <S2/2.
(c) Prove that lim(A: ,_V)_j.(o,o) f(x, y) exists, and find its
value.
2.3 The Derivative
Our goal for this section is to define the derivative of a function f : X C R" — > R" ,
where n and m are arbitrary positive integers. Predictably, the derivative of a
vector-valued function of several variables is a more complicated object than the
derivative of a scalar- valued function of a single variable. In addition, the notion of
differentiability is quite subtle in the case of a function of more than one variable.
We first define the basic computational tool of partial derivatives. After do-
ing so, we can begin to understand differentiability via the geometry of tangent
planes to surfaces. Finally, we generalize these relatively concrete ideas to higher
dimensions.
Partial Derivatives
Recall that ifF:XcR^Risa scalar- valued function of one variable, then the
derivative of F at a number a e X is
F(a + h)-F(a)
F(a)=hm . (1)
fc->o h
Moreover, F is said to be differentiable at a precisely when the limit in equation
(1) exists.
DEFINITION 3.1 Suppose f:X C R" R is a scalar- valued function of n
variables. Let x = (x\ , x%, . . . , xn) denote a point of R" . A partial function
F with respect to the variable Xj is a one-variable function obtained from
/ by holding all variables constant except x, . That is, we set xj equal to a
constant aj for j ^ i. Then the partial function in x, is defined by
F(*i) = ,a2,...,Xi,..., an).
EXAMPLE 1 If/(x, y) = (x2 - y2)/(x2 + y2), then the partial functions with
respect to x are given by
2 2
X — ai
F(x) = f(x,a2) =
x1 + aj
where a2 may be any constant. If, for example, a2 = 0, then the partial function is
x2
F{x) = f(x, 0) = ^ = 1.
2.3 | The Derivative 117
y
Domain of/
/
Domain of F
(restriction
of/)
Figure 2.46 The function / of
Example 1 is defined on R2 -
{(0,0)}, while its partial function F
along y = 0 is defined on the
jc-axis minus the origin.
x
Figure 2.47 Visualizing the
partial derivative f^(a, b).
z
X
Figure 2.48 Visualizing the
partial derivative §r(a, b).
Geometrically, this partial function is nothing more than the restriction of / to
the horizontal line y = 0. Note that since the origin is not in the domain of /, 0
should not be taken to be in the domain of F. (See Figure 2.46.) ♦
Remark In practice, we usually do not go to the notational trouble of explic-
itly replacing the xj 's (j 7^ i) by constants when working with partial functions.
Instead, we make a mental note that the partial function is obtained by allowing
only one variable to vary, while all the other variables are held fixed.
DEFINITION 3.2 The partial derivative of / with respect to x; is the
(ordinary) derivative of the partial function with respect to x, . That is, the
partial derivative with respect to x, is F'(x;), in the notation of Definition
3.1. Standard notations for the partial derivative of / with respect to x,- are
df
~z > DXi /(xi ,...,*„), and fXi (xu x„).
dx,
Symbolically, we have
9/ ,. f(xi,...,xi + h,...,xn)-f(xi,...,xn)
— = hm . (2)
dxi h^o h
By definition, the partial derivative is the (instantaneous) rate of change of /
when all variables, except the specified one, are held fixed. In the case where /
is a (scalar-valued) function of two variables, we can understand
ox
geometrically as the slope at the point (a, b, f(a, b)) of the curve obtained by
intersecting the surface z = fix, y) with the plane y = b, as shown in Figure 2.47.
Similarly,
-^-(a, b)
dy
is the slope at (a, b, f(a, fr))ofthe curve formed by the intersection of z = f(x, y)
and x = a, shown in Figure 2.48.
EXAMPLE 2 For the most part, partial derivatives are quite easy to compute,
once you become adept at treating variables like constants. If
f(x, y) = x2y + cos(x + y),
then we have
— = 2xy - sin(x + y).
dx
(Imagine y to be a constant throughout the differentiation process.) Also
df 7
— = x — sin(x + y).
dy
Chapter 2 | Differentiation in Several Variables
Figure 2.49 The tangent line to
y = F(x) at x = a has equation
y = F(a) + F'(a)(x - a).
(Imagine x to be a constant.) Similarly, if g(x, y) = xy/(x2
quotient rule of ordinary calculus, we have
(x2 + y2)y - xy(2x) y(y2 -
gx(x, y) = -
y2), then, from the
x2)
and
?y(*> y ) =
(x2 + y2)2
(x2 + y2)x — xy(2y)
(x2 + y2)2 '
x(x2 — y2)
(x2 + y2)2 (x2 + y2)2 '
Note that, of course, neither g nor its partial derivatives are defined at (0, 0). ♦
EXAMPLE 3 Occasionally, it is necessary to appeal explicitly to limits to eval-
uate partial derivatives. Suppose /: R2 —> R is defined by
/(*. y)
3x y — y
x2 + y2
if (x, y) # (0, 0)
if(x,y) = (0,0)
Then, for (x, y) ^ (0, 0), we have
9/ 8xy3
3x (x2 + y2)
2\2
and
df
dy
3x — 6x y
2, ,2
(x2 + y2)2
3/ 3/
But what should — (0, 0) and — (0, 0) be? To find out, we return to Definition
dx dy
3.2 of the partial derivatives:
and
df
dy
df
dx
(0, 0)
(0,0)
lim
/(0 + fc.O)
= hm
h-+o h
lim
h^0
f(0, 0 + h)- /(0, 0)
lim
lim -1 =
Tangency and Differentiability
If F: X c R ^ R is a scalar- valued function of one variable, then to have F
differentiable at a number a e X means precisely that the graph of the curve
y = F(x) has a tangent line at the point (a, F(a)). (See Figure 2.49.) Moreover,
this tangent line is given by the equation
y = F(a) + F'(a)(x - a).
(3)
If we define the function H(x) to be F(a) + F'(a)(x — a) (i.e., H(x) is the right
side of equation (3) that gives the equation for the tangent line), then H has two
properties:
1. H(a) = F(a)
2. H'(a) = F'(a).
In other words, the line defined by y = H(x) passes through the point (a, F(a))
and has the same slope at (a, F(a)) as the curve defined by y = F(x). (Hence,
the term "tangent line.")
Now suppose /:lcR2^Risa scalar-valued function of two variables,
where X is open in R2. Then the graph of / is a surface. What should the tangent
plane to the graph of z = f(x, y) at the point (a, b, f(a, b)) be? Geometrically,
2.3 | The Derivative 119
(a,b,f{a,b))
Figure 2.50 The plane tangent
toz = f(x, y) at
(a,b,f(a,b)).
y = b
Figure 2.51 The tangent plane at
(a, b, f(a, b)) contains the lines
tangent to the curves formed by
intersecting the surface
z = f(x, y) by the planes x = a
and y = b.
the situation is as depicted in Figure 2.50. From our earlier observations, we
know that the partial derivative fx(a, b) is the slope of the line tangent at the
point (a, b, f(a, b)) to the curve obtained by intersecting the surface z = f(x, y)
with the plane y = b. (See Figure 2.51.) This means that if we travel along this
tangent line, then for every unit change in the positive x -direction, there's a change
of fx (a , b) units in the z-direction. Hence, by using formula ( 1) of § 1 .2, the tangent
line is given in vector parametric form as
li(0 = (a, b, /(a, b)) + 0, fx(a, b)).
Thus, a vector parallel to this tangent line is
u = i + fx(a,b)k.
Similarly, the partial derivative fy(a, b) is the slope of the line tangent at the point
(a, b, f(a, b)) to the curve obtained by intersecting the surface z = f(x, y) with
the plane x — a. (Again see Figure 2.51.) Consequently, the tangent line is given
by
12(0 = (a, b, f(a, b)) + t(0, 1, fy(a, b)),
so a vector parallel to this tangent line is
v = j + fy(a,b)k.
Both of the aforementioned tangent lines must be contained in the plane tangent
to z = f{x, y) at (a, b, f(a, b)), if one exists. Hence, a vector n normal to the
tangent plane must be perpendicular to both u and v. Therefore, we may take n
to be
n = u x v = -fx(a, b)\- fy(a, b)\ + k.
Now, use equation ( 1 ) of § 1 .5 to find that the equation for the tangent plane — that
is, the plane through (a, b, f(a, b)) with normal n — is
(-fx(a, b), -fy(a, b), 1) • (x - a, y - b, z - f(a, bj) = 0
or, equivalently,
-fx(a, b)(x -a)- fy(a, b)(y -b) + z- f(a, b) = 0.
By rewriting this last equation, we have shown the following result:
THEOREM 3.3 If the graph of z = f(x, y) has a tangent plane at
(a, b, f(a, b)), then that tangent plane has equation
z = f(a, b) + fx(a, b)(x - a) + fy(a, b)(y - b).
(4)
Note that if we define the function h(x, y) to be equal to f(a, b) + fx(a, b)(x —
a) + fy(a, b)(y — b) (i.e., h(x, y) is the right side of equation (4)), then h has the
following properties:
1. h(a,b) = f(a,b)
dh df
ox ax
dh df
and — (a,b)= — (a, b).
dy dy
In other words, h and its partial derivatives agree with those of / at (a, b).
It is tempting to think that the surface z = fix, y) has a tangent plane at
(a, b, f(a, b)) as long as you can make sense of equation (4), that is, as long as the
1 20 Chapter 2 | Differentiation in Several Variables
Figure 2.52 If two points
approach (0, 0, 0) while remaining
on one face of the surface
described in Example 4, the
limiting plane they and (0, 0, 0)
determine is different from the one
determined by letting the two
points approach (0, 0, 0) while
remaining on another face.
partial derivatives fx(a, b) and fy(a, b) exist. Indeed this would be analogous to
the one-variable situation where the existence of the derivative and the existence
of the tangent line mean exactly the same thing. However, it is possible for a
function of two variables to have well-defined partial derivatives (so that equation
(4) makes sense) yet not have a tangent plane.
EXAMPLE 4 Let f(x, y) = \\x\ - \y\\ - \x\ - |y| and consider the surface
defined by the graph of z = f(x, y) shown in Figure 2.52. The partial derivatives
of / at the origin may be calculated from Definition 3.2 as
A(0, 0) = lim
h— >0
/(0 + ft,0)-/(0, 0)
= lim
and
fy(0, 0) = lim
h^0
/(O, 0 + h)- /(O, 0)
= lim
h^0
h
= lim 0 = 0
= lim 0 = 0.
h^0
(Indeed the partial functions F(x) = f(x, 0) and G(y) = f(0, y) are both identi-
cally zero and thus, have zero derivatives.) Consequently, if 'the surface in question
has a tangent plane at the origin, then equation (4) tells us that it has equation
z = 0. But there is no geometric sense in which the surface z = f(x, y) has a
tangent plane at the origin. If we think of a tangent plane as the geometric limit of
planes that pass through the point of tangency and two other "moving" points on
the surface as those two points approach the point of tangency, then Figure 2.52
shows that there is no uniquely determined limiting plane. ♦
Example 4 shows that the existence of a tangent plane to the graph of
z = f(x, y) is a stronger condition than the existence of partial derivatives. It
turns out that such a stronger condition is more useful in that theorems from the
calculus of functions of a single variable carry over to the context of functions
of several variables. What we must do now is find a suitable analytic definition of
differentiability that captures this idea. We begin by looking at the definition of
the one-variable derivative with fresh eyes.
By replacing the quantity a + h by the variable x , the limit equation in formula
(1) may be rewritten as
F(x) - F(a)
F (a) = hm .
This is equivalent to the equation
F(x)
lim
F(a)
F'(a) = 0.
The quantity F'(a) does not depend on x and therefore may be brought inside the
limit. We thus obtain the equation
F(x) - F(a)
lim
F'(a) =0.
Finally, some easy algebra enables us to conclude that the function F is differen-
tiable at a if there is a number F'(a) such that
lim
F(x) - [F(a) + F'(a)(x - a)]
= 0.
(5)
What have we learned from writing equation (5)? Note that the expression in
brackets in the numerator of the limit expression in equation (5) is the function
2.3 | The Derivative 121
(x,F(x))
^\(a, F(a)) \
(x, H(x)^
a x
Figure 2.53 If F is differentiable
at a, the vertical distance between
F(x) and H(x) must approach
zero faster than the horizontal
distance between x and a does.
H(x) that was used to define the tangent line to y
may rewrite equation (5) as
F(x) at (a, F(a)). Thus, we
lim
F{x) - H(x)
= 0.
For the limit above to be zero, we certainly must have that the limit of the numerator
is zero. But since the limit of the denominator is also zero, we can say even more,
namely, that the difference between the y-values of the graph of F and of its tangent
line must approach zero faster than x approaches a. This is what is meant when
we say that "H is a good linear approximation to F near a." (See Figure 2.53.)
Geometrically, it means that, near the point of tangency, the graph of y = F(x)
is approximately straight like the graph of y = H(x).
If we now pass to the case of a scalar- valued function fix , y) of two variables,
then to say that z = fix, y) has a tangent plane at (a, b, f(a, b)) (i.e., that / is
differentiable at ia, b)) should mean that the vertical distance between the graph
of / and the "candidate" tangent plane given by
z = h(x, y) = fia, b) + fx(a, b){x -a)+ fy(a, b\y - b)
must approach zero faster than the point (jc, y) approaches (a,b). (See Fig-
ure 2.54.) In other words, near the point of tangency, the graph of z = fix, y) is
approximately flat just like the graph of z = h(x, y). We can capture this geometric
idea with the following formal definition of differentiability:
DEFINITION 3.4 Let X be open in R2 and /: X C R2 -> R be a scalar-
valued function of two variables. We say that / is differentiable at (a, b) e X
if the partial derivatives fx(a, b) and fy(a, b) exist and if the function
h(x< y) = f(a< b) + fx(fl, b\x -a) + fy(a, b)(y
is a good linear approximation to / near (a, b) — that is, if
fix,y)-h(x,y)
b)
lim
ix,y)-ia,b)\\
= 0.
Moreover, if / is differentiable at (a, b), then the equation z = h(x, y) de-
fines the tangent plane to the graph of / at the point (a, b, fia, b)). If /
is differentiable at all points of its domain, then we simply say that / is
differentiable.
(x,y,f(x,y))
(x,y,h(x,y))
ia,b,fia,b))
Figure 2.54 If / is differentiable at (a, b), the distance
between fix, y) and h(x, y) must approach zero faster than
the distance between (x, y) and (a, b) does.
Chapter 2 | Differentiation in Several Variables
EXAMPLE 5 Let us return to the function f(x, y) = \\x\ - \y\\ - \x\ - \y\ of
Example 4. We already know that the partial derivatives fx (0, 0) and /v(0, 0)
exist and equal zero. Thus, the function h of Definition 3.4 is the zero function.
Consequently, / will be differentiable at (0,0) just in case
f(x,y)-h(x,y) f(x,y)
hm = hm
(a-,y)^(0,0) ||(*, y) - (0, 0)|| C*,y)->-(0,0) ||(x, y)||
\x\-\y\\-\x\-\y\
= lim
(x,v)^(o,o) yx2 +
r
is zero. However, it is not hard to see that the limit in question fails to exist. Along
the line v = 0, we have
f(x,y) IM-01- W-I0I 0 Q
ll(*,y)ll \x
but along the line y = x, we have
f(x,y) = lk|-|x||-|x|-|x| = -2|*| = _^
\\(x,y)\\ s/x2 + x2 </2\x\
Hence, / fails to be differentiable at (0, 0) and has no tangent plane at (0, 0, 0)>
The limit condition in Definition 3.4 can be difficult to apply in practice.
Fortunately, the following result, which we will not prove, simplifies matters in
many instances. Recall from Definition 2.3 that the phrase "a neighborhood of
a point P in a set X" just means an open set containing P and contained in X.
THEOREM 3.5 Suppose X is open in R2. If /: X —> R has continuous partial
derivatives in a neighborhood of (a, b) in X, then / is differentiable at (a, b).
A proof of a more general result (Theorem 3. 10) is provided in the addendum
to this section.
EXAMPLE 6 Let /(*, y) = x2 + 2y2. Then df/dx = 2x and df/dy = Ay,
both of which are continuous functions on all of R2. Thus, Theorem 3.5 implies
that / is differentiable everywhere. The surface z = x2 + 2y2 must therefore
have a tangent plane at every point. At the point (2, — 1), for example, this tangent
plane is given by the equation
z = 6 + 4(x - 2) - 4(y + 1)
(or, equivalently, by Ax — Ay — z = 6). ♦
While we're on the subject of continuity and differentiability, the next result is
the multivariable analogue of a familiar theorem about functions of one variable.
THEOREM 3.6 If /: X C R2 R is differentiable at (a, b), then it is continu-
ous at (a, b).
2.3 | The Derivative 123
EXAMPLE 7 Let the function /: R2
/(*. y) --
2 2
x y
R be denned by
if(jc, 30^(0,0)
x4 + y4
0 if(x,y) = (0,0)
The function / is not continuous at the origin, since lim^ ^j^^o.o) fixi y) does
not exist. (However, / is continuous everywhere else in R .) By Theorem 3.6, /
therefore cannot be differentiate at the origin. Nonetheless, the partial derivatives
of / do exist at the origin, and we have
and
fix, 0) =
7(0, .v) =
o
x4 + 0
0+ v4
0
0
df
dx
df
dy
(0,0) = 0,
(0,0) = 0,
since the partial functions are constant. Thus, we see that if we want something
like Theorem 3.6 to be true, the existence of partial derivatives alone is not
enough. ♦
Differentiability in General
It is not difficult now to see how to generalize Definition 3.4 to three (or more)
variables: For a scalar-valued function of three variables to be differentiable at a
point (a, b, c), we must have that (i) the three partial derivatives exist at (a, b, c)
and (ii) the function h : R3 — > R defined by
h(x, y, z) = f(a, b, c) + fx{a, b, c)(x — a)
+ fy(a, b, c)(y -b) + ft(a, b, c)(z - c)
is a good linear approximation to / near (a, b, c). In other words, (ii) means that
f(x,y,z)-h(x,y,z)
hm = 0.
(x,y,zy+(a,b,c) ||(x, V, z) ~ (a, b, C)\\
The passage from three variables to arbitrarily many is now straightforward.
DEFINITION 3.7 Let X be open in R" and /: X R be a scalar-valued
function; let a = (a\, a%, . . . , a„) e X. We say that / is differentiable at
a if all the partial derivatives fXj(a), i = 1, . . . , n, exist and if the function
h:Rn -> R defined by
H*) = /(a) + Ai(a)(xi - fli) + - a2)
H h fXn{&)(xn - an) (6)
is a good linear approximation to / near a, meaning that
hm = 0.
*->■» II x — all
1 24 Chapter 2 | Differentiation in Several Variables
We can use vector and matrix notation to rewrite things a bit. Define the
gradient of a scalar- valued function /: X C R" R to be the vector
df_ df_ df_
V/(x) =
Consequently,
V/(a) = (A-1(a),/X2(a),..., A„(a)).
Alternatively, we can use matrix notation and define the derivative of / at a,
denoted Df(a), to be the row matrix whose entries are the components of V /(a);
that is,
»/(») = [/*(») M») ■■■ /,„(a)].
Then, by identifying the vector x — a with the n x 1 column matrix whose entries
are the components of x — a, we have
V/(a).(x-a) = D/(a)(x-a) = [/Tl(a) /,2(a) ••• A„(a)]
X\ — d\
x2 - a2
= /xi(a)(*i - fli) + /x2(a)(x2 - a2)
H 1- /v„(a)(x„ - an).
Hence, vector notation allows us to rewrite equation (6) quite compactly as
ft(x)=/(a) + V/(a).(x-a).
Thus, to say that /z is a good linear approximation to / near a in equation (6)
means that
||x - a||
Compare equation (7) with equation (5). Differentiability of functions of one and
several variables should really look very much the same to you. It is worth noting
that the analogues of Theorems 3.5 and 3.6 hold in the case of n variables.
The gradient of a function is an extremely important construction, and we
consider it in greater detail in §2.6.
You may be wondering what, if any, geometry is embedded in this general
notion of differentiability. Recall that the graph of the function /:lcR"->R
is the hypersurface in R"+1 given by the equation xn+\ = f(xi, x2, ... , xn).
(See equation (2) of §2.1.) If / is differentiable at a, then the hypersurface deter-
mined by the graph has a tangent hyperplane at (a, /(a)) given by the equation
x„+i = h(x{ ,x2,...,x„) = /(a) + V/(a) • (x - a)
= /(a) + £>/(a)(x - a). (8)
Compare equation (8) with equation (3) for the tangent line to the curve y = F(x)
at (a, F(a)). Although we cannot visualize the graph of a function of more than
two variables, nonetheless, we can use vector notation to lend real meaning to
tangency in n dimensions.
EXAMPLE 8 Before we drown in a sea of abstraction and generalization, let's
do some concrete computation. An example of an "« -dimensional paraboloid" in
2.3 | The Derivative 1 25
Rn+1 is given by the equation
Xn+\ — X\ + *2 + ■ ■ ■ + X2,
that is, by the graph of the function /(xi, . . . ,x„) = x2 + x| H + x2. We have
— = 2x;- , z = 1, 2, . . . , n,
dxi
so that
V/(*i
x„) = (2xi, 2x2,
2x„).
Note that the partial derivatives of / are continuous everywhere. Hence, the
n -dimensional version of Theorem 3.5 tells us that / is differentiate everywhere.
In particular, / is differentiable at the point (1,2, ...,«),
V/(l,2, ...,n) = (2,4, ...,2n),
and
Df(l,2,...,n)= [2 4 ■■■ In].
Thus, the paraboloid has a tangent hyperplane at the point
(1,2, \2 + 22 + ---+n2)
whose equation is given by equation (8):
x,1+1 =(l2 + 22 +
= (l2 + 22 +
+ n2)+[2 4 ■■■ 2n]
X\ — 1
x2 - 2
+ n2) + 2(xi - 1) + 4(x2 -!) + ■■■+ 2n{xn - n)
= (l2 + 22 +
+ n2) + 2x] + 4x2 + •
(2 ■ 1 + 4 ■ 2 H \-2n-n)
= 2xi + 4x2 H h 2«x„ - (l2 + 22 +
2«x„
+ «2)
= 2i'x,
n(n + l)(2n + 1)
(The formula l2 + 22 + ■ ■ ■ + n2 = n(n + l)(2n + l)/6 is a welhlmown identity,
encountered when you first learned about the definite integral. It's straightforward
to prove using mathematical induction.) ♦
At last we're ready to take a look at differentiability in the most general setting
of all. Let X be open in R" and let f: X —> Rm be a vector- valued function of n
variables. We define the matrix of partial derivatives of f, denoted Df, to be
Chapter 2 | Differentiation in Several Variables
the m x n matrix whose ijth entry is df/dxj, where f:X C R"
component function of f. That is,
R is the (th
Df(x\,x2, ...,xn) =
dxi
dx\
dxi
dX2
d_h
dx2
Mm
dX2
dxn
M.
dx„
Mm
dxn
The ith row of Df is nothing more than Dfj — and the entries of Df are precisely
the components of the gradient vector V f, (Indeed in the case where m = 1,
V/ and Df mean exactly the same thing.)
EXAMPLE 9 Suppose f: R3 -> R2 is given by f(x, y, z) = (x cos y + z, xj).
Then we have
Df(x,y,z) =
cos y —x sin y
y ^
We generalize equation (7) and Definition 3.7 in an obvious way to make the
following definition:
DEFINITION 3.8 (Grand definition of differentiability) Let X c
R" be open, let f : X ->• R", and let a e X. We say that f is differentiable at
a if Df(a) exists and if the function h: R" -> R" defined by
h(x) = f(a) + Df(a)(x - a)
is a good linear approximation to f near a. That is, we must have
lim
II fW - h(x)
lim
|f(x)-[f(a) + Df(a)(x-a)
= 0.
Some remarks are in order. First, the reason for having the vector length
appearing in the numerator in the limit equation in Definition 3.8 is so that there
is a quotient of real numbers of which we can take a limit. (Definition 3 . 7 concerns
scalar-valued functions only, so there is automatically a quotient of real numbers.)
Second the term Df(a)(x — a) in the definition of h should be interpreted as the
product of the m x n matrix Df(a) and the nxl column matrix
x2
Q2
Because of the consistency of our definitions, the following results should
not surprise you:
2.3 | The Derivative 1 27
THEOREM 3.9 If f: X C R" Rm is differentiable at a, then it is continuous
at a.
THEOREM 3.10 If f:lcR"^ R"' is such that, for i = l,...,m and
j = 1, . . . , n, all dfi/dxj exist and are continuous in a neighborhood of a in
X, then f is differentiable at a.
THEOREM 3.1 1 A function f: X C R" -> Rm is differentiable ataeX (in the
sense of Definition 3.8) if and only if each of its component functions fi'.X C
R" — » R, i' = 1, . . . , m, is differentiable at a (in the sense of Definition 3.7).
The proofs of Theorems 3.9, 3.10, and 3.11 are provided in the addendum
to this section. Note that Theorems 3.10 and 3.1 1 frequently make it a straight-
forward matter to check that a function is differentiable: Just look at the partial
derivatives of the component functions and verify that they are continuous. Thus,
in many — but not all — circumstances, we can avoid working directly with the
limit in Definition 3.8.
EXAMPLE 10 The function g: R3 - {(0, 0, 0)} R3 given by
3
g(*> y,z)
x2 + y2 + z2
has
Dg(x, y, z) =
-6z
—6x —6y
(x2 + y2 + z2)2 (x2 + y2 + z2)2 (x2 + y2 + z2)2
y x 0
z 0 x
Each of the entries of this matrix is continuous over R3 — {(0, 0, 0)}. Hence, by
Theorem 3. 10, g is differentiable over its entire domain. ♦
What Is a Derivative?
Although we have defined quite carefully what it means for a function to be
differentiable, the derivative itself has really taken a "backseat" in the preceding
discussion. It is time to get some perspective on the concept of the derivative.
In the case of a (differentiable) scalar-valued function of a single variable,
/:XcR->R, the derivative f'ia) is simply a real number, the slope of the
tangent line to the graph of / at the point (a, f(a)). From a more sophisticated
(and slightly less geometric) point of view, the derivative f'{a) is the number such
that the function
h(x) = /(c) + f'(a)(x - a)
is a good linear approximation to f(x) for x near a. (And, of course, y = h(x) is
the equation of the tangent line.)
If a function /:XcR"-^Rof« variables is differentiable, there must
exist n partial derivatives 3 f/dx\ , . . . , df/dx„ . These partial derivatives form the
components of the gradient vector V/ (or the entries of the 1 x n matrix Df). It
Chapter 2 | Differentiation in Several Variables
is the gradient that should properly be considered to be the derivative of /, but in
the following sense: V/(a) is the vector such that the function h: R" —> R given
by
fc(x)=/(a) + V/(a).(X-a)
is a good linear approximation to /(x) for x near a. Finally, the derivative of a
differentiable vector- valued function f : X c R" — > R"! may be taken to be the
matrix Df of partial derivatives, but in the sense that the function h: R" — > R"1
given by
h(x) = f(a) + Df(a)(x - a)
is a good linear approximation to f(x) near a. You should view the derivative
Df(a) not as a "static" matrix of numbers, but rather as a matrix that defines a
linear mapping from R" to R'". (See Example 5 of §1.6.) This is embodied in
the limit equation of Definition 3.8 and, though a subtle idea, is truly the heart of
differential calculus of several variables.
In fact, we could have approached our discussion of differentiability much
more abstractly right from the beginning. We could have defined a function f:lc
R" — > R°' to be differentiable at a point a 6 X to mean that there exists some
linear mapping L: R" — > R'" such that
Hm ||f(x)-[f(a) + L(x-a)]|| = Q
x^a ||x — a||
Recall that any linear mapping L: R" — > R'" is really nothing more than multipli-
cation by a suitable m x n matrix A (i.e., that L(y) = Ay). It is possible to show
that if there is a linear mapping that satisfies the aforementioned limit equation,
then the matrix A that defines it is both uniquely determined and is precisely the
matrix of partial derivatives Df(a). (See Exercises 60-62 where these facts are
proved.) However, to begin with such a definition, though equivalent to Definition
3.8, strikes us as less well motivated than the approach we have taken. Hence, we
have presented the notions of differentiability and the derivative from what we
hope is a somewhat more concrete and geometric perspective.
Addendum: Proofs of Theorems 3.9, 3.10, and 3.11
Proof of Theorem 3.9 We begin by claiming the following: Let xeR" and
B = (bjj) be an m x n matrix. If y = Bx, (so y 6 Rm), then
llyll <TOI, (9)
/ ,\l/2
where K = ( >_\ . bf- 1 . We postpone the proof of (9) until we establish the
main theorem.
To show that f is continuous at a, we will show that ||f(x) — f(a)|| —> 0 as
x -> a. We do so by using the fact that f is differentiable at a (Definition 3.8).
We have
||f(x) - f(a)|| = ||f(x) - f(a) - Df(a)(x - a) + Df(a)(x - a)||
< ||f(x) - f(a) - £>f(a)(x - a)|| + ||Df(a)(x - a)||, (10)
using the triangle inequality. Note that the first term in the right side of inequality
(10) is the numerator of the limit expression in Definition 3.8. Thus, since f is
2.3 | The Derivative 1 29
differentiable at a, we can make ||f(x) — f(a) — £>f(a)(x — a)|| as small as we wish
by keeping ||x — a|| appropriately small. In particular,
||f(x)-f(a)-Df(a)(x-a)|| < ||x-a||
if || x — a|| is sufficiently small. To the second term in the right side of inequality
(10), we may apply (9), since Df(a) is an m x n matrix. Therefore, we see that if
|| x — a|| is made sufficiently small,
||f(x)-f(a)|| < ||x-a||+Z||x-a||=(l + Z)||x-a||. (11)
The constant K does not depend on x. Thus, as x —> a, we have
||f(x) - f(a)|| -» 0,
as desired.
To complete the proof, we establish inequality (9). Writing out the matrix
multiplication,
y = Bx =
b\\%\ + ^12^2 H V bXnxn
bi\x\ + b22x2 H h b2nx„
bm\X\ + bm2X2 + ■■■ + &
bi • x
b2 -x
where b, denotes the ith row of B, considered as a vector in R". Therefore, using
the Cauchy-Schwarz inequality,
||y|| = ((bi • X)2 + (b2 • X)2 + • • • + (bm • X)2)
< (||b1||2||x||2 + ||b2|r||x|r +
,2\ 1/2
+ l|bm||2||x||2)1/2
|b1||2 + ||b2||2 + --- + ||b„
2)'
Now,
Consequently,
= bjl+bf2 + ---+bl = J2bfJ.
llbi ||2 + ||b2 1|2 + • • • + ||b„, ||2 = lib, II2 = E E bl = Rl-
i=i i=i j=\
Thus, ||y|| < AT||x||, and we have completed the proof of Theorem 3.9. ■
Proof of Theorem 3.10 First, we prove Theorem 3. 10 for the case where / is a
scalar-valued function of two variables. We begin by writing
f(x\ , x2) — f(flu a2) = f(x\ , xi) - f{a\ , x2) + f(a\ , x2) - f(a\, a2).
By the mean value theorem,2 there exists a number c\ between a\ and xi such
that
f{x\,x2) - f(ai,x2) = fXi(c\, x2){xx - ai)
2 Recall that the mean value theorem says that if F is continuous on the closed interval [a , b] and differen-
tiable on the open interval (a, b), then there is a number c in (a , b) such that F(b) — F(a) = F'(c)(b — a).
Chapter 2 | Differentiation in Several Variables
and a number c2 between a2 and x2 such that
f(ai,x2) - f(aua2) = fX2(au c2)(x2 - a2).
(This works because in each case we hold all the variables in / constant except
one, so that the mean value theorem applies.) Hence,
\f(xux2) - f(a\, a2) - fxi{a\, a2){x\ - a{) - fX2(aua2)(x2 - a2)\
= \fXl(c\,x2)(x\ - ai)+ fX2(ai, c2)(x2 - a2) - fX](au a2)(x\ - ax)
-fX2(au a2)(x2 - a2)\
< x2)(.x;i - ai) - fXl(ax,a2)(xi - fli)|
+ |/x2(ai, c2)(x2 - a2) - fXl(ai, a2)(x2 - a2)\ ,
by the triangle inequality. Hence,
\f(xi,x2) - f(aua2) - fxi(ai,a2)(xi - a\) - fX2(au a2)(x2 - a2)\
< |/*i(Cl,*2) ~ fxM\'a2)\ \x\ ~ a\\
+ \fX2(au c2) - fX2{ax, a2)\ \x2 - a2\
< {\fXl(ci,x2) - fXl(ai,a2)\ + \fX2(ai, c2) - fX2(au a2)\) ||x - a||,
since, for i = 1,2,
\Xi -en\< ||x - a|| = ((xi - ax)2 + (x2 - a2)2)1'2.
Thus,
|/(xi, x2) - f(au a2) - fxi(aua2)(xi - a{) - fX2{ax,a2){x2 - a2)\
l|x-a||
< \fxi(c\,x2)- fxl(ai,a2)\ + \fX2(auc2)- fX2(ai,a2)\. (12)
As x — > a, we must have that c, —> a\ , for i = I, 2, since c, is between a, and x, .
Consequently, by the continuity of the partial derivatives, both terms of the right
side of (12) approach zero. Therefore,
,. \f(xi,x2)- f(ax,a2)- fXl(ai,a2)(xi - a{) - fX2(a{ , a2)(x2 - a2)\
hm 1 = 0
||x-a||
as desired.
Exactly the same kind of argument may be used in the case that / is a scalar-
valued function of n variables — the details are only slightly more involved so
we omit them. Granting this, we consider the case of a vector-valued function
f: R" — > Rm. According to Definition 3.8, we must show that
||ml|fW-f(a)-Z>fW(x-a)|=0
||x - a||
The component functions of the expression appearing in the numerator may be
written as
G, = Mx) - fi(z) - D/,(a)(x - a), (14)
where fi, i = 1, . . . , m, denotes the ith component function of f. (Note that, by
the cases of Theorem 3.10 already established, each scalar- valued function /, is
2.3 I Exercises
differentiable.) Now, we consider
||f(x) - f(a) - Df(a)(x - a)|| \\(GU G2, . . . , G„
|x-a|| ||x -a||
_(G] + Gl + --- + Gjy/2
Hi -a||
< |Gil + |G2| + --- + |Gm|
l|x-a||
Igil ] \G2\ | | \G„
lix — a|| ||x — a|| ||x-a||
Asx^ a, eachterm |G, |/||x — a|| -> 0, by definition of G, in equation (14) and
the differentiability of the component functions f\ of f . Hence, equation (13) holds
and f is differentiable at a. (To see that (G2 H h G2 )1/2 < \G\\-\ hi Gm \ ,
note that
(IGH H h |Gm|)2 = |Gi|2 H h |G,„|2
+ 2|Gi| |G2| +2|Gi| |G3| + ■ ■ ■ + 2|GM_i| |G„,|
> |Gi|2 + --- + |GJ2.
Then, taking square roots provides the inequality.) ■
Proof of Theorem 3.11 In the final paragraph of the proof of Theorem 3. 10, we
showed that
||f(x)-f(a)-Pf(a)(x-a)|| < |Gt| | |G2| | | |G,„|
||x-a|| llx — a|| ||x-a|| l|x-a||
where G, = fi(x) — /,(a) — Z)/}(a)(x — a) as in equation (14). From this, it fol-
lows immediately that differentiability of the component functions f\ , . . . , fm at
a implies differentiability of fat a. Conversely, for i = 1, . . . , m,
||f(x)-f(a)-Df(a)(x-a)|| = \\(GU G2, . . . , Gm)|| > |G,|
||x-a|| ||x-a|| ||x-a||
Hence, differentiability of f at a forces differentiability of each component
function. ■
2.3 Exercises
In Exercises 1—9, calculate df/dx and df/dy.
1- f(x,y) = xy2+x2y
2. f(x, y) = e^1
3. f(x, y) = sinxy + cosjcy
6. f(x, y) = \n(x2 + y2)
7. f(x, y) = cosx3y
' x '
8. f(x, y) = In
4. f(x,y) .
5. f{x,y) .
^ - y2
\+x2 + 3/
x2-y2
x2 + y2
9. f(x, y) = xey + y sin(x2 + y)
In Exercises 10—17, evaluate the partial derivatives dF/dx,
dF/dy, and dF/dz for the given functions F.
10. F(x,y,z) = x + 3y-2z
Chapter 2 | Differentiation in Several Variables
x — v
11. F(x,y,z)= — ~
y + z
12. F(x, y, z) = xyz
13. F(x,y,z) = y/x2 + y2 + z2
14. F(x, y, z) = eax cosfcy + eaz sinbx
-ic T7i ^ x + y + z
15. F(x, y, z) = — —
(1 + x2 + y2 + Z2)3/2
16. F(x, y, z) = sinx2y3z4
17. F(x,y,z)
x3 + yz
X2 + Z2 + 1
FzW //ze gradient V/(a), where f and a are given in Exer-
cises 18-25.
18. /(x, y) = x2y + ey'x, a = (1,0)
x — y
19. f(x,y) .
-, a = (2,-1)
.v2 + y2 + V
20. /(x, y, z) = sinxyz, a = (jt, 0, jr/2)
21. /(x, j, z) = xy + y cosz — * sinyz,
a = (2,-l,jr)
22. f(x, y) = c»> + In (x - y), a = (2, 1)
23. /(x,y,z)=^±2, a = (3, -1,0)
24. /(x,y,z) = coszln(x + y2), a = (e, 0, jt/4)
2 _ 2
25. /(x,y,z) = / Z , ■ = (-1,2,1)
yz + zz + 1
In Exercises 26-33, find the matrix Df( a) of partial derivatives,
where f and a are as indicated.
26. f(x,y)=-, a = (3, 2)
y
27. f(x,y,z) = x2+xln(yz), a = (-3, e,e)
28. f(x, y, z) = (2x - 3y + 5z, x2 + y, In (yz)),
a = (3, -1,-2)
29. f(x, y, z) = (xyz, ^x2 + y2 + z2^),
a = (1,0, -2)
30. f(f) = (r, cos2f, sin 5/), a = 0
31. f(x, y, z, id) = (3x - 7y + z, 5x + 2z - Sw,
y-\7z + 3w), a = (1,2, 3, 4)
32. f(x, y) = (x2y, x + y2, cos jrxy), a = (2, -1)
33. f(s, r) = (j2,jf,r2), a = (-1,1)
Explain why each of the functions given in Exercises 34—36 is
differentiable at every point in its domain.
34. /(x, y) = xy — 7x8y2 + cosx
x + v + z
35. /(x,y,z):
36. f(x, y) :
xy
xl + y4 y
37. (a) Explain why the graph of z = x3 — 7xy + ey has
a tangent plane at (— 1 , 0, 0).
(b) Give an equation for this tangent plane.
38. Find an equation for the plane tangent to the graph of
z = 4cosxy at the point (tt/3, 1, 2).
39. Find an equation for the plane tangent to the graph of
z = ex+y cosxy at the point (0, 1, e).
40. Find equations for the planes tangent to z =
x2 — 6x + y3 that are parallel to the plane
4x - 12y + j
7.
41. Use formula (8) to find an equation for the hy-
perplane tangent to the 4-dimensional paraboloid
x5 = 10 — (x\ + 3x| + 2x| + x|) at the point
(2,-1, 1,3,-8).
42. Suppose that you have the following information con-
cerning a differentiable function /:
/(2, 3) =12, /(1.98,3) = 12.1, f(2, 3.01) = 12.2.
( a) Give an approximate equation for the plane tangent
to the graph of / at (2, 3, 12).
(b) Use the result of part (a) to estimate /(l .98, 2.98).
In Exercises 43—45, (a) use the linear function h(\) in Def-
inition 3.8 to approximate the indicated value of the given
function f. (b) How accurate is the approximation determined
in part (a)?
43. f(x, y) = ex+y, /(0.1, -0.1)
44. f(x, y) = 3 + cos itxy, /(0.98, 0.51)
45. f(x, y, z) = x2 + xyz + y3z, /(1.01, 1.95, 2.2)
46. Calculate the partial derivatives of
xi + Xj H 1- x„
/(Xl,X2,
47. Let
fix, y)
xy
, x„)
■x2y + 3x3
*? + xl + ■
■+x,
x2 + y2
0
if(x,y)^(0, 0)
if(x,y) = (0,0)
x2 + y2 + z2
(a) Calculate df/dx and df/dy for (x, y)/(0, 0).
(You may wish to use a computer algebra system
for this part.)
(b) Find fx(0, 0) and /,(0, 0).
As mentioned in the text, if a function F{x) of a single variable
is differentiable at a, then, as we zoom in on the point {a, F(a)),
the graph of y = F(x) will "straighten out" and look like its
tangent line at (a, F(a)). For the differentiable functions given
2.3 I Exercises 133
in Exercises 48-51, (a) calculate the tangent line at the indi-
cated point, and (b) use a computer to graph the function and
the tangent line on the same set of axes. Zoom in on the point
of tangency to illustrate how the graph of y = F(x) looks like
its tangent line near (a, F(a)).
^ 48. F(x)
2x + 3, a = 1
smx, a
^ 49. F{x) = x + -
>-3 - 3x2 + x
x —ox -(- X
O 50. F(x) = T— , a = 0
xL + 1
^ 51. F(x) = ln(x2 + 1), a = -1
52. (a) Use a computer to graph the function F(x) =
(x-2)2'\
(b) By zooming in near x = 2, offer a geometric dis-
cussion concerning the differentiability of F at
x =2.
As discussed in the text, a function fix, y) may have partial
derivatives fx(a, b) and fy(a, b) yet fail to be differentiable at
(a, b). Geometrically, if a function fix, y) is differentiable at
(a, b), then, aswezoom in on thepoint(a, b, f(a, b)), thegraph
of z = fix, y) will "flatten out" and look like the plane given
by equation (4) in this section. For the functions fix, y) given
in Exercises 53-57, (a) calculate fx(a, b) and fy(a, b) at the
indicated point (a , b) and write the equation for the plane given
by formula (4) of this section, (b) use a computer to graph the
equation z = f(x, y) together with the plane calculated in part
(a). Zoom in near the point (a, b, f(a, b)) and discuss whether
or not f(x, y) is differentiable at (a, b). (c) Give an analytic
(i.e., nongraphical) argument for your answer in part (b).
ft 53. f(x, y) = x3-xy + y2, (a, b) = (2, 1)
^> 54. f{x, y) = Hx - \)yf'\ (a, b) = (1, 0)
^55. fix,y) .
^56. fix,y) .
xy
(fl,fc) = (0,0)
x2 + y2 + 1 '
fit 3tt \
sinxcosy, (a, &)=(—,— )
V 6 4 /
^ 57. fix,y) = x2 s'my + y2cosx, (a,A)=^— ,— )
58. Let gix, y) = ^fxy.
(a) Is g continuous at (0, 0)?
(b) Calculate dg/dx and dg/dy whenxy ^ 0.
(c) Show that gx(0, 0) and g, (0, 0) exist by supplying
values for them.
(d) Are dg/dx and dg/dy continuous at (0, 0)?
(e) Does the graph of z = gix, y) have a tangent plane
at (0, 0)? You might consider creating a graph of
this surface.
(f) Is g differentiable at (0, 0)?
59. Suppose f: R" —> W is a linear mapping; that is,
f(x) = Ax, where x = (xi, X2, . . . , xn) e R"
and A is an m x n matrix. Calculate Df(x) and relate
your result to the derivative of the one-variable linear
function fix) = ax.
In Exercises 60-62 you will establish that the matrix -Df(a) of
partial derivatives of the component functions ofiis uniquely
determined by the limit equation in Definition 3.8.
60. Let X be an open set in R", let aeX, and let F:ic
R" -± R". Show that
lim ||F(x)|| = 0
lim F(x) = 0.
61. Let X be an open set in R", let ael, and let f:XC
R" — >• R" . Suppose that A and B are m x n matrices
such that
lim
x— >a
lim
||f(x)-[f(a) + A(x-a)]|
llx — a||
f(x) - [f(a) + 5(x - a)] ||
x^a || x — a||
(a) Use Exercise 60 to show that
hmCS-AKx-a,
x->-a Hx — all
0.
(b) Write x — a as fh, where h is a nonzero vector in
R" . First argue that
lim
x-»a
(S - A)(x - a)
0 implies
lim(B-A)W
t-o \\th\\
and then use this result to conclude that A = B.
(Hint: Break into cases where t > 0 and where
t < 0.)
62. Let X be an open set in R", let aeX, and let f:XC
R" — >• R" . Suppose that A is an m x n matrix such
that
lim
x->a
||f(x)-[f(a) + A(x-a)]
0.
In this problem you will establish that A = Df(a).
(a) Define F:ICR"-> R" by
F(x) =
f(x) - f(a) - A(x - a)
Identify the ;th component function us-
ing component functions of f and parts of the
matrix A.
(b) Note that under the assumptions of this problem
and Exercise 60, we have that limx^a F(x) = 0.
1 34 Chapter 2 | Differentiation in Several Variables
First argue that, for i=l,...,m, we have
limx^a Fi(x) = 0. Next, argue that
lim F,(x) = 0 implies lim F,-(a + he:) = 0,
where denotes the standard basis vector
(0, 1, ...,0)forR".
(c) Use parts (a) and (b) to show that atj
M.
(a),
where a,; denotes the ijth entry of A. (Hint: Break
into cases where h > 0 and where h < 0.)
2.4 Properties; Higher-order Partial
Derivatives
Properties of the Derivative
From our work in the previous section, we know that the derivative of a function
f : X c R" R'" can be identified with its matrix of partial derivatives. We next
note several properties that the derivative must satisfy. The proofs of these results
involve Definition 3.8 of the derivative, properties of ordinary differentiation, and
matrix algebra.
PROPOSITION 4.1 (Linearity of differentiation) Let f,g:Jfc R" -» Rffl
be two functions that are both differentiable at a point aeX, and let c e R be
any scalar. Then
1. The function h = f + g is also differentiable at a, and we have
Z)h(a) = D(f + g)(a) = Df(a) + £»g(a).
2. The function k = cf is differentiable at a and
£>k(a) = £>(cf)(a) = c£>f(a).
EXAMPLE 1 Let f and g be defined by f(x, y) = (x + y, xy siny, y/x) and
g(x, y) = (x2 + y2, yexy , 2x3 — 7y5). We have
Df(x, y) =
1
y siny
L -3V*2
and
Dg(x,y) =
2x
y2exy
6x2
1
x siny + xy cosy
l/x
2y
ex>' + xye^^
-35y4
Thus, by Theorem 3.10, f is differentiable on R2 — {y-axis} and g is differentiable
on all of R2. If we let h = f + g, then part 1 of Proposition 4. 1 tells us that h must
be differentiable on all of its domain, and
Dh(x,y) = Df(x,y) + Dg(x,y)
2x + \ 2y + 1
y sin y + y2exy x sin y + xy cos y + exy + xyexy
6x2 -y/x2 l/x- 35y4
2.4 | Properties; Higher-order Partial Derivatives
Note also that the function k = 3g must be differentiable everywhere by part 2
of Proposition 4.1. We can readily check that Dk(x, y) = 3Dg(x, y): We have
k(x, y) = (3x2 + 3y2, 3yexy , 6x3 - 21y5).
Hence,
£>k(x, y) =
6.v
6y
3exy + 3xye
18x2
-105y4
2x
2y
y2exy
exy + xyexy
6x2
-35y4
= 3
= 3£>g(x, y). 4
Due to the nature of matrix multiplication, general versions of the product
and quotient rules do not exist in any particularly simple form. However, for
scalar-valued functions, it is possible to prove the following:
PROPOSITION 4.2 Let /,g:lcR"^Rbe differentiable ataeX Then
1. The product function fg is also differentiable at a, and
D(fgM = s(a)D/(a) + /(a)Z)g(a).
2. If g(a) / 0, then the quotient function f/g is differentiable at a, and
g(a)D/(a) - /(a)Z)g(a)
D(//g)(a) =
g(a)2
(/#)(*> y, z) = (xyz + 2yz2 - xz2)exy,
EXAMPLE 2 If f(x, y, z) = z<?x-v and g(x, y, z) = xy + 2yz - xz, then
so that
D(fg)(x,y,z)
(yz - z2)exy + (xyz + 2yz2 - xz2)yexy
(xz + 2z2)exy + (xyz + 2yz2 - xz2)xexy
(xy + 4yz — 2xz)ex>'
-i t
Also, we have
and
Df(x,y,z)= [yzexy xZexy exy]
Dg(x,y,z)=[y - z x + 2z 2y-x],
so that
g(x, y, z)Df(x, y, z) + f(x, y, z)Dg(x, y, z)
(xy2z + 2y2z2 — xyz2)exy
(x2yz + 2xyz2 - x2z2)exy
(xy + 2yz - xz)exy
(yz - z2)exy
(xz + 2z2)exv
(2yz - xz)exy
-1 T
Chapter 2 | Differentiation in Several Variables
= e
xy2z
x2yz -
- 2y2z2 - xyz2 + yz- z2
2xyz2 - x2z2 + xz + 2z2
xy + 4yz — 2xz
which checks with part 1 of Proposition 4.2. (Note: The matrix transpose is used
simply to conserve space on the page.) ♦
The product rule in part 1 of Proposition 4.2 is not the most general
result possible. Indeed, if /:ICR"->R is a scalar- valued function and
g:lcR"^ R'" is a vector- valued function, then if / and g are both differ-
entiate at a e X, so is fg, and the following formula holds (where we view g(a)
as an m x 1 matrix):
£>(/g)(a) = g(a)D/(a) + /(a)Dg(a).
Partial Derivatives of Higher Order
Thus far in our study of differentiation, we have been concerned only with partial
derivatives of first order. Nonetheless, it is easy to imagine computing second-
and third-order partials by iterating the process of differentiating with respect to
one variable, while all others are held constant.
EXAMPLE 3 Let f(x, y, z) = x2y + y2z. Then the first-order partial deriva-
tives are
9/ „ 9/ 2 „ 9/2
— = 2xy, — = x + 2yz, and — = y ,
ax ay az
The second-order partial derivative with respect to x, denoted by d2f/dx2 or
fxx(x, y, z), is
d2f 9 (df\ 3 .„ , „
— — = — — = — (2xy) = 2y-
dx2 dx \dx ) dx
Similarly, the second-order partials with respect to y and z are, respectively,
92/ 9 /3A d
dy2 dy \dy J dy^
and
d2f 3 (df
dz2 dz \ dz J dz ^
There are more second-order partials, however. The mixed partial derivative
with respect to first x and then y, denoted d2 f /dydx or fxy(x, y, z), is
92/ 9 /3/\ 3 ,
= — (2xy) = 2x.
dydx dy \dx ) dy
There are five more mixed partials for this particular function: d2f/dxdy,
d2f/dzdx, d2 f /dxdz, d2 f /dzdy, and d2 f /dydz. Compute each of them to get
a feeling for the process. ♦
In general, if /: X C R" R is a (scalar- valued) function of n variables, the
&th-order partial derivative with respect to the variables (in that
2.4 | Properties; Higher-order Partial Derivatives
order), where ii , 1*2, -..,£* are integers in the set {1, 2, .... n} (possibly repeated),
is the iterated derivative
dxh- ■ ■ dxhdxil dxk dxkdxil
Equivalent (and frequently more manageable) notation for this /rth-order partial
is
Note that the order in which we write the variables with respect to which we
differentiate is different in the two notations: In the subscript notation, we write
the differentiation variables from left to right in the order we differentiate, while
in the 3 -notation, we write those variables in the opposite order (i.e., from right
to left).
EXAMPLE 4 Let f(x, y, z, w) = xyz + xy2w — cos(x + zw). We then have
d2f 9 9 ,
fyw(x,y,z,w)= — — = — — (xyz + xy w - cos(x + zw))
away aw ay
9
= — (xz + 2xyw) = 2xy,
dw
and
d2f 9 9 2
fwy(x, y, z, w) = —— = — —(xyz + xy w - cos(x + zw))
ayaw ay aw
9
9^
(xy2 + z sin(x + zw)) = 2xy.
♦
Although it is generally ill-advised to formulate conjectures based on a single
piece of evidence, Example 4 suggests that there might be an outrageously simple
relationship among the mixed second partials. Indeed, such is the case, as the next
result, due to the 1 8th-century French mathematician Alexis Clairaut, indicates.
THEOREM 4.3 Suppose that X is open in R" and /:XcR"^R has con-
tinuous first- and second-order partial derivatives. Then the order in which we
evaluate the mixed second-order partials is immaterial; that is, if i\ and 12 are any
two integers between 1 and n, then
J2J_ = J2J_
dxj1 dxj2 dxi2 dxj1
A proof of Theorem 4.3 is provided in the addendum to this section. We
also suggest a second proof (using integrals!) in Exercise 4 of the Miscellaneous
Exercises for Chapter 5.
It is natural to speculate about the possibility of an analogue to Theorem 4.3
for &th-order mixed partials. Before we state what should be an easily anticipated
result, we need some terminology.
Chapter 2 | Differentiation in Several Variables
DEFINITION 4.4 Assume X is open in R". A scalar-valued function
/:XcR"^R whose partial derivatives up to (and including) order at
least k exist and are continuous on X is said to be of class Ck. If / has
continuous partial derivatives of all orders on X, then / is said to be of class
C°°, or smooth. A vector-valued function f: X c R" -> R'" is of class Ck
(respectively, of class C°°) if and only if each of its component functions is
of class Ck (respectively, C°°).
THEOREM 4.5 Let /: X c R" -> R be a scalar-valued function of class Ck.
Then the order in which we calculate any £th-order partial derivative does not
matter: If . . . , /<-) are any k integers (not necessarily distinct) between 1 and
n, and if (ji, ... , jk) is any permutation (rearrangement) of these integers, then
dkf = dkf
dxj{ ■ ■ ■ dxjt dxjj ■ ■ ■ dxjt
EXAMPLE 5 If f(x, y, z, w) = x2weyz — zexw + xyzw, then you can check
that
95f d5f
2eyz(yz + \)
dxdwdzdydx dzdydwd2x'
verifying Theorem 4.5 in this case.
Addendum: Two Technical Proofs
Proof of Part 1 of Proposition 4.1
Step 1. We show that the matrix of partial derivatives of h is the sum of
those of f and g. If we write h(x) as (hi(x), /i2(x), . . . , hm(x)) (i.e., in terms of
its component functions), then the ijth entry of £>h(a) is dhj/dxj evaluated at a.
But hj(x) = fi(x) + gi(x) by definition of h. Hence,
dh; 9 dfj dg;
— ^ = —(/;■« + oo) = -f- +
dXj dXj axj oxj
by properties of ordinary differentiation (since all variables except xj are held
constant). Thus,
^— (a)= — (a)+ — (a),
axj axj axj
and therefore,
Z)h(a) = £>f(a) + Dg(a).
Step 2. Now that we know the desired matrix of partials exists, we must
show that h really is differentiable; that is, we must establish that
Hm ||h(x)-[h(a) + Ph(a)(x-a)]|| = Q
2.4 | Properties; Higher-order Partial Derivatives
(a, b + Ay)
(a + Ax, b + Ay)
•
•
+
+
•
(a'b)
(a + Ax, b)
Figure 2.55 To construct the
difference function D used in the
proof of Theorem 4.3, evaluate /
at the four points shown with the
signs as indicated.
As preliminary background, we note that
||h(x)-[h(a) + /)h(a)(x-a)]||
l|x-a||
||f(x) + g(x) - [f(a) + g(a) + £>f(a)(x - a) + Dg(a)(x - a)]||
II* -a||
||(f(x) - [f(a) + Df(a)(x - a)]) + (g(x) - [g(a) + Dg(a)(x - a)])||
II* -a||
< ||f(x)-[f(a) + Df(a)(x-a)]|| | ||g(x) - [g(a) + Dg(a)(x - a)]||
x-a
x - a
by the triangle inequality, formula (2) of §1.6. To show that the desired limit
equation for h follows from the definition of the limit, we must show that given
any e > 0, we can find a number S > 0 such that
.,n , , . t, ||h(x)-[h(a) + £>h(a)(x-a)]||
if 0 < || x - a|| < 8, then < e.
x - a
(1)
Since f is given to be differentiable at a, this means that given any e\ > 0, we can
find S\ > 0 such that
•fn ,i ,i * a llf(x)-[f(a) + flf(a)(x-a)]|
if 0 < || x — a|| < Si, then < ei.
x-a
(2)
Similarly, differentiability of g means that given any 62 > 0, we can find a 82 > 0
such that
Tn „ „ . t, ||g(x)-[g(a) + JDg(a)(x-a)]|
if 0 < || x — a|| < h, then < €2.
x - a
(3)
Now we're ready to establish statement (1). Suppose e > 0 is given. Let 81
and 82 be such that (2) and (3) hold with ei = €2 = e/2. Take 8 to be the smaller
of 81 and 82- Hence, if 0 < ||x — a|| < <5, then both statements (2) and (3) hold
(with €1 = €2 = e/2) and, moreover,
||h(x)-[h(a) + Ph(a)(x-a)]|| ||f(x) - [f(a) + Df(a)(x - a)]||
x-a
Hi -a||
|g(x)-[g(a) + Z)g(a)(x-a)]|
+
x-a
<el+e2
e e
= 2 + 2=€-
That is, statement (1) holds, as desired. ■
Proof of Theorem 4.3 For simplicity of notation only, we'll assume that / is a
function of just two variables (x and y). Let the point (a , b) e R2 be in the interior
of some rectangle on which fx, fy, fxx, fyy, fxy, and fyx are all continuous.
Consider the following "difference function." (See Figure 2.55.)
D(Ax, Ay) = f(a + Ax, b + Ay) - f(a + Ax, b)
-f(a,b+ Ay) + f(a,b).
1 40 Chapter 2 | Differentiation in Several Variables
(a, b + Ay) (a + Ax, b + Ay)
R
{'d
X
(a,b)
(a + Ax, b)
Figure 2.56 Applying the mean
value theorem twice.
Our proof depends upon viewing this function in two ways. We first regard D as
a difference of vertical differences in /:
D(Ax, Ay) = [f(a + Ax,b + Ay) - f{a + Ax, b)]
- [f(a, b + Ay) - f (a, b)]
= F(a + Ax) - F(a).
Here we define the one-variable function F(x) to be f(x, b + Ay) — f(x, b). As
we will see, the mixed second partial of / can be found from two applications of
the mean value theorem of one-variable calculus. Since / has continuous partials,
it is differentiable. (See Theorem 3. 10.) Hence, F is continuous and differentiable,
and, thus, the mean value theorem implies that there is some number c between
a and a + Ax such that
D(Ax, Ay) = F(a + Ax) - F(a) = F'(c)Ax.
(4)
Now F'{c) = fx{c, b + Ay) — fx(c, b). We again apply the mean value theorem,
this time to the function fx(c, y). (Here, we think of c as constant and y as the
variable.) By hypothesis fx is differentiable since its partial derivatives, fxx and
fxy, are assumed to be continuous. Consequently, the mean value theorem applies
to give us a number d between b and b + Ay such that
F'(c) = fx(c, b + A.y) - fx(c, b) = fxy(c, d)Ay.
Using equation (5) in equation (4), we have
D(Ax, Ay) = F'(c)Ax = fxy(c, d)AyAx.
(5)
The point (c, d) lies somewhere in the interior of the rectangle R with vertices
(a, b), {a + Ax, b), (a, b + Ay), (a + Ax, b + Ay), as shown in Figure 2.56.
Thus, as (Ax, Ay) -> (0, 0), we have (c, d) — > (a, b). Hence, it follows that
fxy(c, d) -> fxy(a, b) as (Ax, Ay)
since fxy is assumed to be continuous. Therefore,
(0, 0),
D(Ax, Ay)
fxJa,b)= lim fxJc,d)= lim .
(Ajc,A)0-»-(o,O) " (Ax,Ajo^(0,0) AyAx
On the other hand, we could just as well have written D as a difference of
horizontal differences in /:
£>(Ax, Ay) = [/(a + Ax, b + Ay) - f(a, b + Ay)]
- [f(a + Ax, b) - f(a, b)]
= G(b + Ay) - G(b).
Here G(y) = f(a + Ax, y) — f(a, y). As before, we can apply the mean value
theorem twice to find that there must be another point (c, d) in R such that
D(Ax, Ay) = G'(d)Ay = fyx(c, d)AxAy.
Therefore,
fyx(a, b) =
lim fvx(c, d)
(Ax,Av)^(0,0) "
lim
(Ajt,Ay)->-(0,0)
D(Ax, Ay)
Ax Ay
Because this is the same limit as that for fxy(a, b) just given, we have established
the desired result. ■
2.4 I Exercises 141
2.4 Exercises
In Exercises 1—4, verify the sum rule for derivative matrices
(i.e., part 1 of Proposition 4.1) for each of the given pairs of
functions:
1. f(x, y) — xy + cosx, g(x, y) = sin(xy) + y3
2. f{x, y) = (ex+y,xey), g(x, y) = Qn(xy), yex)
3. f(x, y, z) = (x sin y + z, yez — 3x2), g(x, y, z) =
(x3 cosx, xyz)
4. f(x, y, z) = (xyz2, xe~y , y sinxz), g(x, y, z) =
(x — y, x2 + y2 + z2, ln(xz + 2))
Verify the product and quotient rules (Proposition 4.2) for the
pairs of functions given in Exercises 5—8.
x
5- f(x, y) = x2y + y3, g(x, y) = -
y
6. f(x,y) = exy, g(x, y) = x sin2y
7- fix, y) = 3xy + y5 , g(x, y) = x3 - 2xy2
8. f(x,y,z) = xcos(yz),
g(x, y, z) = x2 + x9y2 + y2z3 + 2
For the functions given in Exercises 9-1 7 determine all second-
order partial derivatives (including mixed partials).
9.
fix, y) =
x3y7 + 3xy2 — Ixy
10.
fix, y) =
cos (xy)
11.
fix, y) =
ey/x _ ye-x
12.
fix, y) =
siny7*2 + y2
13.
fix, y) =
1
sin2 x + 2ey
14.
fix,y) =
15.
fix, y) =
y sin x — x cos y
16.
fix, y) =
17.
fix,y) =
x2ey + e2z
18.
fix, y, z)
x — y
y + z
19.
f(x,y,z)
= x2yz + xy2z + xyz2
20.
fix, y, z)
= exyz
21.
fix, y, z)
= eax sin y + ehx cos z
22.
Consider
the function F(x,y,z)
y3Z5 — Ixyz.
(a) Find Fxx, Fyy, and F-z.
(b) Calculate the mixed second-order partials Fxy,
Fyx, Fxz, F^x, Fyz, and Fzy, and verify Theorem
4.3.
(c) Is Fxyx = Fxxyl Could you have known this with-
out resorting to calculation?
(d) Is Fxyz = Fyzx7
23. Let f(x, y) = ye3x. Give general formulas for
d"f/dxn and dnf/dyn, where n > 2.
24. Let f(x,y,z) = xe2y + yeiz+ze~x. Give general
formulas for d"f/dx", dnf/dyn, and d"f/dz", where
n > 1.
25. Let f(x, y, z) = In ^— ^. Give general formulas for
d"f/dx", d"f/dyn, and dnf/dz", where n > 1. What
can you say about the mixed partial derivatives?
26. Let f(x, y, z) = x1y2z3 - 2x4yz.
(a) What is d4f/dx2dydzl
(b) What is dsf/dx*dydzl
(c) What is 915//9*133v9z?
27. Recall from §2.2 that a polynomial in two variables x
and y is an expression of the form
d
p(x, y)=Yl ckixky',
k.l=0
where c« can be any real number for 0 < k, I < d. The
degree of the term cuxky[ when cu ^ 0 is k + I and
the degree of the polynomial p is the largest degree
of any nonzero term of the polynomial (i.e., the largest
degree of any term for which cu / 0). For example,
the polynomial
p(x, y) = 7*y + 2x2y3 - 3x4 - 5xy3 + 1
has five terms of degrees 15, 5, 4, 4, and 0. The de-
gree of p is therefore 1 5 . (Note: The degree of the zero
polynomial p(x, y) = 0 is undefined.)
(a) If p(x, y) = 8jc7v10 — 9x2y + 2x, what is the de-
gree of dp/dx7 dp/dyi d2p/dx27 32p/dy2?
d2p/dxdy?
(b) If p(x, y) = Sx2y + 2x3y, what is the degree of
dp/dx? dp/dyl d2p/dx27 d2p/dy2? d2p/dxdy?
(c) Try to formulate and prove a conjecture relating
the degree of a polynomial p to the degree of its
partial derivatives.
28. The partial differential equation
32/ 3*7 32/
— - H H = 0
dx2 dy2 dz2
is known as Laplace's equation, after Pierre Simon
de Laplace (1749-1827). Any function / of class C2
1 42 Chapter 2 | Differentiation in Several Variables
that satisfies Laplace's equation is called a harmonic
function.3
(a) Is f(x, y, z) = x2 + y2 — 2z2 harmonic? What
about f(x, y, z) = x2 - y2 + z21
(b) We may generalize Laplace's equation to functions
of n variables as
32/ 32/ 92/ „
— y + — T + ■■■+— V=0.
ox{ 9xj 9jc„
Give an example of a harmonic function of n vari-
ables, and verify that your example is correct.
29. The three-dimensional heat equation is the partial dif-
ferential equation
/d2T d2T d2T\ _ dT
\~d~x2 + ~d~y2 + Ik2) ~ ~dt'
where k is a positive constant. It models the tempera-
ture T(x, y, z, t) at the point (x, y, z) and time t of a
body in space.
(a) We examine a simplified version of the heat equa-
tion. Consider a straight wire "coordinatized" by
x. Then the temperature T(x, t) at time t and po-
sition x along the wire is modeled by the one-
dimensional heat equation
d2T _ dT
Jx2 ~ ~dt'
Show that the function T(x, t) = e~kt cosjc satis-
fies this equation. Note that if t is held constant at
value ?o, then T(x, to) shows how the temperature
varies along the wire at time to. Graph the curves
z = T(x, to) for to = 0,1, 10, and use them to un-
derstand the graph of the surface z = T(x,t) for
t > 0. Explain what happens to the temperature of
the wire after a long period of time.
(b) Show that T(x, y, t) = e~k'(cosx + cos y) satis-
fies the two-dimensional heat equation
id2T d2T\ _ dT
[jx* + Jy2) = It'
Graph the surfaces given by z = T(x, y, to), where
to = 0, 1, 10. If we view the function T(x, y, t) as
modeling the temperature at points (x, y) of a flat
plate at time t, then describe what happens to the
temperature of the plate after a long period of time.
(c) Now show that T(x, y, z, t) = e~kt(cosx +
cos y + cos z) satisfies the three-dimensional heat
equation.
30. Let
[-^(4x4) 30^(0,0)
/(je>y)=J \x2 + y2J
[ 0 if(x,y) = (0,0)
(a) Find fx (x,y) and /,(* , y) for (x, y) ? (0, 0). (You
will find a computer algebra system helpful.)
(b) Either by hand (using limits) or by means of
part (a), find the partial derivatives /r(0, y) and
/,(x,0).
(c) Find the values of fxy(0, 0) and /y.,-(0, 0). Recon-
cile your answer with Theorem 4.3.
A surface that has the least surface area among all surfaces
with a given boundary is called a minimal surface. Soap bub-
bles are naturally occurring examples of minimal surfaces. It
is a fact that minimal surfaces having equations of the form
z = fix, y) (where f is of class C2) satisfy the partial differ-
ential equation
(1 + l]) ZXX + (1 + Zl) Zyy = 2ZxZyZxy. (6)
Exercises 31—33 concern minimal surfaces and equation (6).
31 . Show that a plane is a minimal surface.
32. Scherk's surface is given by the equation ez cos y =
cosx.
(a) Use a computer to graph a portion of this surface.
(b) Verify that Scherk's surface is a minimal surface.
33. One way to describe the surface known as the helicoid
is by the equation x = y tan z.
(a) Use a computer to graph a portion of this surface.
(b) Verify that the helicoid is a minimal surface.
2.5 The Chain Rule
Among the various properties that the derivative satisfies, one that stands alone
in both its usefulness and its subtlety is the derivative's behavior with respect
to composition of functions. This behavior is described by a formula known as
3 Laplace did fundamental and far-reaching work in both mathematical physics and probability theory.
Laplace's equation and harmonic functions are part of the field of potential theory, a subject that Laplace
can be credited as having developed. Potential theory has applications to such areas as gravitation, elec-
tricity and magnetism, and fluid mechanics, to name a few.
2.5 | The Chain Rule 143
the chain rule. In this section, we review the chain rule of one- variable calculus
and see how it generalizes to the cases of scalar- and vector-valued functions of
several variables.
The Chain Rule for Functions of One Variable: A Review
We begin with a typical example of the use of the chain rule from single-variable
calculus.
EXAMPLE 1 Let f(x) = sinx and x(t) = t3 + t. We may then construct the
composite function /(x(r)) = sin(f3 + t). The chain rule tells us how to find the
derivative of/oi with respect to t :
(/ o jc)'(r) = -^-(sin(73 + 0) = (cos(?3 + t))(3t2 + 1).
at
Since x = r3 + t, we have
(/ o x)'(t) = isinjc) ■ ^{t1 + 0 = f{x) ■ x'(t). ♦
ax at
In general, suppose X and T are open subsets of R and /:XcR^ R
and x : T C R — >• R are functions defined so that the composite function
/ o x : T -> R makes sense. (See Figure 2.57.) In particular, this means that the
range of the function x must be contained in X, the domain of /. The key result
is the following:
/
-i ) ( — h
X
R R R
Figure 2.57 The range of the function x must be contained in the domain X of / in
order for the composite / o x to be defined.
THEOREM 5.1 (The chain rule in one variable) Under the preceding as-
sumptions, if x is differentiable at to 6 Tand /isdifferentiableat.To = x(to) 6 X,
then the composite / o x is differentiable at to and moreover,
(/ o x)'(to) = f\xo)x'(to). (1)
A more common way to write the chain rule formula in Theorem 5.1 is
df df dx
Mt0) = Mx0) —(to). (2)
dt dx dt
Although equation (2) is most useful in practice, it does represent an unfor-
tunate abuse of notation in that the symbol / is used to denote both a function
of x and one of t. It would be more appropriate to define a new function y by
y(t) = (f o x)(t) sothatdy/df = (df/dx)(dx/dt). But our original abuse of no-
tation is actually a convenient one, since it avoids the awkwardness of having too
many variable names appearing in a single discussion. In the name of simplicity,
we will therefore continue to commit such abuses and urge you to do likewise.
The formulas in equations (1) and (2) are so simple that little more needs
to be said. We elaborate, nonetheless, because this will prove helpful when we
1 44 Chapter 2 | Differentiation in Several Variables
generalize to the case of several variables. The chain rule tells us the following:
To understand how / depends on t, we must know how / depends on the "in-
termediate variable" x and how this intermediate variable depends on the "final"
independent variable t. The diagram in Figure 2.58 traces the hierarchy of the
variable dependences. The "paths" indicate the derivatives involved in the chain
rule formula.
Figure 2.58 The chain rule for functions of a single variable.
Figure 2.59 The composite function fox.
The Chain Rule in Several Variables
Now let's go a step further and assume /:XcR2^RisaC' function of
two variables and x: T c R -> R2 is a differentiable vector-valued function
of a single variable. If the range of x is contained in X, then the composite
/ox:TcR^Ris defined. (See Figure 2.59.) It's good to think of x as
describing a parametrized curve in R2 and / as a sort of "temperature func-
tion" on X. The composite / o x is then nothing more than the restriction of / to
the curve (i.e., the function that measures the temperature along just the curve).
The question is, how does / depend on t ? We claim the following:
PROPOSITION 5.2 Suppose x: T C R -> R2 is differentiable at t0 e T, and
/: X C R2 —> R is differentiable at xo = x(?o) = (xq, yo) £ X, where T and X
are open in R and R2, respectively, and range x is contained in X. If, in addition,
/ is of class C1, then fox:T^ R is differentiable at to and
df df dx df dy
dt dx dt dy dt
Before we prove Proposition 5.2, some remarks are in order. First, notice
the mixture of ordinary and partial derivatives appearing in the formula for the
2.5 | The Chain Rule 145
derivative. These terms make sense if we contruct an appropriate "variable hier-
archy" diagram, as shown in Figure 2.60. At the intermediate level, / depends
on two variables, x and y (or, equivalently, on the vector variable x = (x , y)),
so partial derivatives are in order. On the final or composite level, / depends on
just a single independent variable t and, hence, the use of the ordinary derivative
df /dt is warranted. Second, the formula in Proposition 5.2 is a generalization of
equation (2): A product term appears for each of the two intermediate variables.
Figure 2.61 The graph of the
function x of Example 2.
EXAMPLE 2 Suppose f(x, y) = (x + y2)/(2x2 + 1) is a temperature func-
tion on R2 and x(?) = (2t, t + 1). The function x gives parametric equations
for a line. (See Figure 2.61.) Then
(/ox)(0 = /(x(0)
2t + (t + iy t2 + At + i
8r2 + 1 8?2 + 1
is the temperature function along the line, and we have
df A- lAt - 32t2
dt
(St2 + l)2
by the quotient rule. Thus, all the hypotheses of Proposition 5.2 are satisfied and
so the derivative formula must hold. Indeed, we have
and
Therefore,
dx
df
dy
At)
\-2x2
Axy2
(2x2 + l)2
2y
2x2 + r
dx dy
dt ' dt
(2, 1).
df dx df dy
dx dt dy dt
1 - 2x2 - Axy2
(2x2 + l)2
2(1 - St2 - 8t(t
2 +
2v
2jc2 + 1
(8f2 + l)2
}f) + 2(t
1)
8f2 + 1
1 46 Chapter 2 | Differentiation in Several Variables
after substitution of 2t for x and t + 1 for y. Hence,
df dx df dy _ 2(2 -It- \6t2)
dx~dt + ~dy~dt ~ (8?2 + l)2 '
which checks with our previous result for df/dt. ♦
Proof of Proposition 5.2 Denote the composite function / o x by z- We want to
establish a formula for dz/dt at to- Since z is just a scalar- valued function of one
variable, differentiability and the existence of the derivative mean the same thing.
Thus, we consider
dz z(t)-z(t0)
—(to) = km — - — ,
dt t^t0 t — to
and see if this limit exists. We have
dz,, f(x(t), y(t)) - f(x(t0), y(t0))
—(to) = bm .
dt f-»-«o t — to
The first step is to rewrite the numerator of the limit expression by subtracting
and adding f(x0l y) and to apply a modicum of algebra. Thus,
dzr^ v f(x' - /(*o» y) + f(xo, y) - f(x0, y0)
—(to) = bm
dt t->tQ t — to
,. fix, y) - f(x0, y) f(xo, y) - f(x0, y0)
= hm h bm .
t-*h t — to t-*t<, t — to
(Remember that x(to) = xo = (xo, yo)-) Now, for the main innovation of the proof.
We apply the mean value theorem to the partial functions of /. This tells us that
there must be a number c between x0 and x and another number d between y0
and y such that
and
Thus,
f(x, y) - f(x0, y) = fx(c, y)(x - x0)
f(x0, y) - f(x0, yo) = fy(x0, d)(y - y0).
dz x — xq y — yo
—(to) = hm fx(c, y)- — — + hm fy(x0, d)- — —
dt t->t0 t — to ■ t — to
x(t)-x(to) ,,y(t)-y(to)
= hm fx(c, y) — h hm fy(x0, d) —
t-*k t - 10 t^-to t - to
dx dy
= fx(XQ, yo)-J-(to) + fy(XQ, yo)-T-(t0),
dt dt
by the definition of the derivatives
dx dy
-(to) and -(t0)
and the fact that fx(c, y) and fy(xo, d) must approach fx(xo, yo) and fy(xo, yo),
respectively, as t approaches to, by continuity of the partials. (Recall that / was
assumed to be of class C 1 . ) This completes the proof. ■
Proposition 5.2 and its proof are easy to generalize to the case where /
is a function of n variables (i.e., /:XC R" —> R) and x : T C R — >• R". The
2.5 | The Chain Rule 147
appropriate chain rule formula in this case is
df , , df , . dxi , df , ,dx2, s
= -^(x0)— ^0) + -^(x0)— i(r0) +
at ax\ at 0x2 dt
df ,dx„
+ w~ (xo)-r-(fo)-
3x„ a/
(3)
Note that the right side of equation (3) can also be written by using matrix notation
so that
dj_
dt
Co) =
9/
3xi
(xo)
Thus, we have shown
3/
3x2
(xo)
dx„
(xo)
~dl
dt
(to)
(to)
(to)
df
dt
(t0) = Df(x0)Dx(t0) = V/(x0) -x'(/0),
(4)
where we use x'(fo) as a notational alternative to Dx(fo)- The version of the chain
rule given in formula (4) is particularly important and will be used a number of
times in our subsequent work.
Let us consider further instances of composition of functions of many
variables. For example, suppose X is open in R3, T is open in R2, and
/:XcR3^R and x: T c R2 R3 are such that the range of x is contained
in X. Then the composite / os:rcR2->R can be formed as shown in
Figure 2.62. Note that the range of x, that is, x(7), is just a surface in R3, so
fox can be thought of as an appropriate "temperature function" restricted to this
surface. If we use x = (x, y, z) to denote the vector variable in R3 and t = (s,t)
for the vector variable in R2, then we can write a plausible chain rule formula
from an appropriate variable hierarchy diagram. (See Figure 2.63.) Thus, it is
X
1 48 Chapter 2 | Differentiation in Several Variables
Figure 2.63 The chain rule for / o x, where /:XCR!^R and
\:T c R2 ->- R3.
reasonable to expect that the following formulas hold:
df = dfdx + dfdy dfdz
ds dx ds dy ds dz ds
and (5)
df _ df dx df dy df dz
~dl ~ dx~dl + ~dyJt + JzJt'
(Again, we abuse notation by writing both df/ds, df/dt and df/dx, df/dy,
df/dz.) Indeed, when / is a function of x, y, and z of class C1, formula (3)
with n = 3 applies once we realize that dx/ds, dx/dt, etc., represent ordinary
differentiation of the partial functions in s or t.
EXAMPLE 3 Suppose
f(x, y, z) = x2 + y2 + z2 and x(s, t) = (s cos t, est , s2 - t2).
Then h(s, t) = f o x(s, t) = s2 cos2 1 + e2st + (s2 - t2)2, so that
dh d(f ox) ^ 2 ~ , , 7x
= 2* cos2 f + 2fe2sr + 4s(s2 - t2)
dt dt
We also have
ds ds
= -2s1 cos t sin t + 2seIst - 4t(sz - tl).
dh d(fox) 2 ,, 2
and
9/ - 3/ - 3/ -
— = 2x, — = 2y, — = 2z
dx dy dz
dx dx
— = cosr, — = — jsin/,
ds dt
t st dy st
— = te , — = se ,
ds dt
3z dz
— =2s, — = -It.
ds dt
2.5 | The Chain Rule 149
Hence, we compute
df = d(f o x) = df dx | df dy | df dz
ds ds dx ds dy ds dz ds
= 2x(cos 0 + 2y(test) + 2z(2s)
= 2s cos r(cos 0 + 2est(test) + 2{s2 - t2)(2s)
= 2s cos2 t + 2te2s1 + 4s(s2 - t2),
just as we saw earlier. We leave it to you to use the chain rule to calculate df/dt
in a similar manner. ♦
Of course, there is no need for us to stop here. Suppose we have an open set
X in R'", an open set T in R", and functions /: X —> R and x: T — »• R'" such
that A = / o x: I ^> R can be defined. If / is of class C1 and x is differentiate,
then, from the previous remarks, h must also be differentiate and, moreover,
dh df dxi df 3x2 3/ dxm
dtj dx\ dtj dx2 dtj dxm dtj
= E
7 = 1,2,
df dxk
k=l dxk dtj
Since the component functions of a vector- valued function are just scalar- valued
functions, we can say even more. Suppose f: X c Rm -> RP and x: T c R"
R"! are such that h = f o x: T c R" — >• R'' can be defined. (As always, we assume
that X is open in Rm and T is open in R" .) See Figure 2.64 for a representation of
the situation. If f is of class C 1 and x is differentiable, then the composite h = f o x
is differentiable and the following general formula holds:
dhi
3^
df dxk
dxk dtJ
1,2,
7 = 1,2,
, n.
(6)
The plausibility of formula (6) is immediate, given the variable hierarchy diagram
shown in Figure 2.65.
X
x(T)
R" Rm Rp
Figure 2.64 The composite fo x where f : X C R"' ^ R'' and x: T C R" Rm.
Now comes the real "magic." Recall that if A is a p x m matrix and B is an
m x n matrix, then the product matrix C = AB is defined and is a /? x « matrix.
Moreover, the i/th entry of C is given by
kj-
k=l
Chapter 2 | Differentiation in Several Variables
If we recall that the ijth entry of the matrix Dh(t) is dhi/dtj, and similarly for
Z)f(x) and Dx(t), then we see that formula (6) expresses nothing more than the
following equation of matrices:
Dh(t) = D(f o x)(t) = Df(x)Dx(t). (7)
The similarity between formulas (7) and (1) is striking. One of the reasons
(perhaps the principal reason) for defining matrix multiplication as we have is
precisely so that the chain rule in several variables can have the elegant appearance
that it has in formula (7).
EXAMPLE 4 Suppose f:R3 -> R2 is given by f(xi,x2,x3) =
(jci — x2, xix2x3) and x: R2 -> R3 is given by x(*i, t2) = (ht2, t2
f o x: R2 -> R2 is given by (f o x)(*i , t2) = (tit2 - tf, f3f23), so that
ls 4). Then
D(fox)(t)
On the other hand
Df(X) :
so that the product matrix is
Df(x)Dx(t)
t2 - 2fi
3t24
h
3ty2
1 -1 0
X2Xj, X\Xt, X\X2
h ~ 2*i
and Dx(t)
h *i
2*! 0
0 2*2
X2X-x,t2 + 2x 1X3*1 X2*3fi + 2x\X2t2
*2
t2t3
'1 '2
2*1
2f2
,3
1 '2
2t\t\
after substituting forxi, x2, andx3. Thus, D(f o x)(t) = Df(x)Dx(t), as expected.
Alternatively, we may use the variable hierarchy diagram shown in Figure 2.66
and compute any individual partial derivative we may desire. For example,
dfl _ df^dx^ dfidx^ df2 9x3
3*i 9xi 3*i 3x2 3?i 3x3 3*i
2.5 | The Chain Rule 151
Figure 2.66 The variable hierarchy diagram for Example 4.
by formula (6). Then by abuse of notation,
3/2
= (*2*3)(fc) + (*l*3)(2fi) + (XIX2)(0)
= (t2t22)(t2) + (ht2)(t22)(2h)
which is indeed the (2,1) entry of the matrix product. ♦
At last we state the most general version of the chain rule from a technical
standpoint; a proof may be found in the addendum to this section.
THEOREM 5.3 (The chain rule) Suppose X c R"! and T c R" are open and
f: X — > Kp and x: T — »• R'" are defined so that range x c X. If x is differen-
tiable at t0 6 T and f is differentiable at x0 = x(t0), then the composite f o x is
differentiable at to, and we have
D(fox)(t0)= Df(x0)Dx(t0).
The advantage of Theorem 5.3 over the earlier versions of the chain rule we
have been discussing is that it requires f only to be differentiable at the point in
question, not to be of class C1. Note that, of course, Theorem 5.3 includes all
the special cases of the chain rule we have previously discussed. In particular,
Theorem 5.3 includes the important case of formula (4).
EXAMPLE 5 Let f: R2 -> R2 be defined by f(x, y) = (x - 2y + 7, 3xy2).
Suppose that g: R3 -> R2 is differentiable at (0, 0, 0) and we know that
g(0, 0, 0) = (-2, 1) and
2 4 5 "
Dg(0,0,0)= _i o 1 ■
We use this information to determine D(f o g)(0, 0, 0).
First, note that Theorem 5.3 tells us that fog must be differentiable at (0, 0, 0)
and, second, that
D(f o g)(0, 0, 0) = Df (g0, 0, 0)) Dg(0, 0, 0) = Df(-2, l)Dg(0, 0, 0).
Chapter 2 | Differentiation in Several Variables
Since we know f completely, it is easy to compute that
Df(x,y) =
Thus,
D(fog)(0,0,0)= 3
We remark that we needed the full strength of Theorem 5.3, as we do not know
anything about the differentiability of g other than at the point (0, 0, 0). ♦
1 -2
3y2 6xy
so that Df(-2, 1)
-2
■12
2 4 5
-1 0 1
4 3
12 3
EXAMPLE 6 (Polar/rectangular conversions) Recall that in §1.7 we pro-
vided the basic equations relating polar and rectangular coordinates:
x = r cos 9
y = r sin 9
Now suppose you have an equation defining a quantity w as a function of x and
y; that is,
w = f{x, y).
Then, of course, w may just as well be regarded as a function of r and 9 by
susbtituting r cos 9 for x and r sin 9 for y • That is,
g(r, 0) = f(x(r, 9), y(r, 9)).
Our question is as follows: Assuming all functions involved are differentiable,
how are the partial derivatives dw/dr, dw/d9 related to dw/dx, dw/dy?
In the situation just described, we have w = g(r, 9) = (/ o x)(r, 9), so that
the chain rule implies
Dg(r,9) = Df(x,y)Dx(r,9).
Therefore,
dg
8r
9g
36
dx
df
dx
df
dy
df
dy
dx
dx
W
dy
dy
dr
d9 _
cos 9
—r sin
sin 9
r cos
By extracting entries, we see that the various partial derivatives of w are related
by the following formulas:
3w
"97
dw
~d~9
cost
dw dw
h sm9 —
dx dy
dw dw
-r smw h r zos9 —
dx dy
(8)
The significance of (8) is that it provides us with a relation of differential
operators:
2.5 | The Chain Rule 153
a a
— = cost? —
dr dx
sinO
dy
d a a
— = —r sin 6 h r cos 6 —
dd dx dy
(9)
The appropriate interpretation for (9) is the following: Differentiation with
respect to the polar coordinate r is the same as a certain combination of differen-
tiation with respect to both Cartesian coordinates x and y (namely, the combina-
tion cos 0 d/dx + sm6 d/dy). A similar comment applies to differentiation with
respect to the polar coordinate 0 . Note that, when r ^ 0, we can solve algebraically
for d/dx and d/dy in (9), obtaining
a
— = cosf
dx
sin 6
d
dr
/•
dQ
a
— = sinfi
dy
a
Yr +
cos 6
a
r
d9
(10)
We will have occasion to use the relations in (9) and (10), and the method of
their derivation, later in this text. ♦
Addendum: Proof of Theorem 5.3
We begin by noting that the derivative matrices £>f(x0) and Dx(t0) both exist
because f is assumed to be differentiable at x0 and x is assumed to be differentiable
at to. Thus, the product matrix Df(xo)£>x(to) exists. We need to show that the limit
in Definition 3.8 is satisfied by this product matrix, that is, that
||(fox)(t)-[(fox)(t0) + Df(xo)Dx(t0)(t-to)]||
hm = 0. (11)
||t -toll
In view of the uniqueness of the derivative matrix, it then automatically follows
that f o x is differentiable at to and that Z)f(xo)£>x(to) = D(f o x)(to). Thus, we
entirely concern ourselves with establishing the limit (11) above.
Consider the numerator of (1 1). First, we rewrite
(f o x)(t) - [(f ox)(to) + Df(x0)Dx(t0)(t - t0)]
= (f o x)(t) - (f o x)(t0) - £>f(x0)(x(t) - x(f0))
+ Df(x0)(x(t) - x(f0)) - Df(xo)Dx(t0)(t - t0).
Then we use the triangle inequality:
|| (f o x)(t) - [(f o x)(t0) + £>f(x0)Dx(to)(t - t0)] ||
< || (f o x)(t) - (f o x)(t0) - Df(x0)(x(t) - x(t0))\\
+ ||£»f(x0)(x(t) - x(f0)) - Df(x0)Dx(t0)(t - t0) ||
= || (f o x)(t) - (f o x)(t0) - Df(x0)(x(t) - x(?0))||
+ ||Df(xo) [(x(t) - x(f0)) - Dx(t0)(t - t0)] || .
1 54 Chapter 2 | Differentiation in Several Variables
By inequality (9) in the proof of Theorem 3.9, there is a constant K such that, for
any vector h e R", ||Df(x0)h|| < A"||h||. Thus,
|| (fo x)(t) - (f o x)(t0) - £>f(x0)Dx(t0)(t - t0) ||
< || (f o x)(t) - (f o x)(t0) - Df(x0)(x(t) - x(t0))\\ (12)
+ K||x(t)-x(fo)-JDx(t0Xt-to)||.
To establish the limit (11) formally, we must show that given any e > 0, we
may find a S > 0 such that if 0 < ||t — toll < 8, then
|| (f o x)(t) - [(f o x)(t0) + £>f(xo)Z)x(to)(t - t0)] || <f
lit -toll
Consider the first term of the right side of (12). Using the differentiability of x at
t0 and inequality (1 1) in the proof of Theorem 3.9, we can find some <50 > 0 and
a constant A"o such that if 0 < ||t — toll < So, then
||x(t) - x(t0)|| < K0 ||t -to||.
By the differentiability of f at xo, given any e\ > 0, we may find some Si > 0
such that if 0 < ||x — Xo|| < <$i, then
||f(x)-[f(xo) + Df(x0)(x-xo)] ||
< ei-
l|x-xo||
Setei = e/(2A:0). Withx = x(t),x0 = x(t0), we have that if both 0 < ||t- t0|| <
So and 0 < ||t — t0|| < 8\/Kq, then
||x(t) - x(t0)|| < Ko lit -toll <<5i.
Hence,
||(f o x)(t) - (f o x)(t0) - Z)f(x0)(x(t) - x(t0))|| < ei||x(t) - x(t0)||
< ei^ollt- toll = . (13)
Now look at the second term of the right side of (12). Since x is differentiable
at t0, given any €2 > 0, we may find some S2 > 0 such that if 0 < ||t — t0 1| < S2,
then
||x(t) - [x(t0) + £>x(t0)(t - t0)] ||
lit -toll
Set e2 = e/(2K). Then, for 0 < ||t - t0|| < S2, we have
||x(t)-[x(to) + Dx(to)(t-t0)]|| < 2^llt-t0||. (14)
Finally, let S be the smallest of <5o, <5i/A"o, and 82- Then, for 0 < ||t — to || < 8, we
have that both the inequalities (13) and (14) hold and thus (12) becomes
|| (fo x)(t) - (f o x)(t0) - Df(x0)£»x(t0)(t - t0) ||
< €-\\t - toll +^(^11*- toll)
= e||t- toll-
Hence,
||(f o x)(t) - (f o x)(t0) - £)f(xo)Dx(to)(t - t0)"
lit -toll
as desired.
2.5 I Exercises 155
2.5 Exercises
1. If f(x, y, z) = x2 — y3 + xyz, and x = 6t + 7, y =
sin2r, z = t2, verify the chain rule by finding df /dt
in two different ways.
months from now will be
71 1
s2 + t2,
find
2. If f(x, y) = sin(;ty) and x = s + t, y
df/ds and df/dt in two ways:
(a) by substitution.
(b) by means of the chain rule.
3. Suppose that a bird flies along the helical curve
x = 2 cos t, y = 2 sin t,z = 3f . The bird suddenly en-
counters a weather front so that the barometric pres-
sure is varying rather wildly from point to point as
P(x, y, z) = 6x2z/y atm.
(a) Use the chain rule to determine how the pressure
is changing at t = rc/4 min.
(b) Check your result in part (a) by direct substitution.
(c) What is the approximate pressure at t = rc/4 +
0.01 min?
4. Suppose that z
+ y , where x = st and y is a
function ofs and t . Suppose further that when (s, t) =
dz
(2, 1), dy/dt = 0. Determine —(2, 1).
5. You are the proud new owner of an Acme Deluxe Bread
Kneading Machine, which you are using for the first
time today. Suppose that at noon the dimensions of your
(nearly rectangular) loaf of bread dough are L = 7 in
(length), W = 5 in (width), and H = 4 in (height). At
that time, you place the loaf in the machine for knead-
ing and the machine begins by stretching the loaf's
length at an initial rate of 0.75 in/min, punching down
the loaf's height at a rate of 1 in/min, and increasing
the loaf's width at a rate of 0.5 in/min. What is the rate
of change of the volume of the loaf when the machine
starts? Is the dough increasing or decreasing in size at
that moment?
6. A rectangular stick of butter is placed in the microwave
oven to melt. When the butter's length is 6 in and its
square cross section measures 1 .5 in on a side, its length
is decreasing at a rate of 0.25 in/min and its cross-
sectional edge is decreasing at a rate of 0.125 in/min.
How fast is the butter melting (i.e., at what rate is the
solid volume of butter turning to liquid) at that instant?
7. Suppose that the following function is used to model
the monthly demand for bicycles:
P(x, y) = 200 + 20V0.1* + 10 - lltfy.
In this formula, x represents the price (in dollars
per gallon) of automobile gasoline and y repre-
sents the selling price (in dollars) of each bicycle.
Furthermore, suppose that the price of gasoline t
x = 1 + O.lf — cos ■
b
and the price of each bicycle will be
TCt
y = 200 + 2t sin — .
6
At what rate will the monthly demand for bicycles be
changing six months from now?
8. The Centers for Disease Control and Prevention pro-
vides information on the body mass index (BMI)
to give a more meaningful assessment of a person's
weight. The BMI is given by the formula
BMI :
10,000u;
where w is an individual's mass in kilograms and h
the person's height in centimeters. While monitoring a
child's growth, you estimate that at the time he turned
1 0 years old, his height showed a growth rate of 0.6 cm
per month. At the same time, his mass showed a growth
rate of 0.4 kg per month. Suppose that he was 140 cm
tall and weighed 33 kg on his tenth birthday.
(a) At what rate is his BMI changing on his tenth
birthday?
(b) The BMI of a typical 10-year-old male increases
at an average rate of 0.04 BMI points per month.
Should you be concerned about the child's weight
gain?
9. A cement mixer is pouring concrete in a conical pile.
At the time when the height and base radius of the
concrete cone are, respectively, 30 cm and 12 cm, the
rate at which the height is increasing is 1 cm/min and
the rate at which the volume of cement in the pile is
increasing is 320 cm3/min. At that moment, how fast
is the radius of the cone changing?
1 0. A clarinetist is playing the glissando at the beginning of
Rhapsody in Blue, while Hermione (who arrived late)
is walking toward her seat. If the (changing) frequency
of the note is / and Hermione is moving toward the
clarinetist at speed v, then she actually hears the fre-
quency (j> given by
c + v
f,
where c is the (constant) speed of sound in air, about
330 m/sec. At this particular moment, the frequency is
/ = 440 Hz and is increasing at a rate of 100 Hz per
second. At that same moment, Hermione is moving
toward the clarinetist at 4 m/sec and decelerating at
Chapter 2 | Differentiation in Several Variables
2 m/sec2. What is the perceived frequency cf> she
hears at that moment? How fast is it changing? Does
Hermione hear the clarinet's note becoming higher or
lower?
11. Suppose z = f(x,y) has continuous partial deriva-
tives. Let x = e'cos8, y = e'smd. Show that
then
+
-2r
+
dz.
X z
1 8. Suppose that ui = g I — , — I is a differentiable func-
\y yJ
tion of u = x/y and v = z/y. Show then that
dw dw dw
xir + yir + zir=0-
ox ay dz
In Exercises 19-27, calculate D(io g) in two ways: (a) by first
evaluating fog and (b) by using the chain rule and the deriva-
tive matrices Df and Dg.
12. Suppose that z = f(x, y) has continuous partial
derivatives. Let x = 2uv and y = u2 + v2 . Show that
then
dz dz
2x
du dv
13. If w = g(u2
+
, dz dz
+ 4v .
dx 3y
u2) has continuous partial
derivatives with respect to x = u2 — v2 and y = v2 —
u2, show that
dw dw
v h u — =0.
du dv
14. Suppose that z = f(x + y, x — y) has continuous par-
tial derivatives with respect to u = x + y and v =
x — y. Show that
dz
dz
(du)
(dz
dx
dy "
\dv
15. If w = f
xy
xy
x2 + y
is a differentiable function of
- , show that
x2 + y2
dw dw
x h y = 0.
dx dy
(x2-y2\
16. If w = f —z r- is a differentiable function of
\x2 + y2J
x2 - y2
u = — , show that then
xl + y1
dw dw
x h y = 0.
dx dy
I y — x z — x \
1 7. Suppose w = f [ , ) is a differentiable
V xy xz
y — x z — x
function of u = and v = . Show then that
xy
xz
r. dw i dw i dw
x ir + y IT + z IT = °-
dx dy dz
19. f(x) = (3x5, e2x), g(s, t) = s- It
20. f(x) = (x2, cosSx, Inx), g(s, t,u) = s + t2 + i
21. f(x, y) = yex, g(s, t) = (s - t, s + t)
22. f(x, y)
23. f(x, y) =
3y2,g(s,t) = (st,s + t2)
(xv-^ + v3),g(,,r)=(^2r)
y x
x' y
24. f(x, y, z) = (x2y + y2z, xyz, ez),
g(t) = (f - 2, 3f + 7, f3)
25. f(x, y) = (xy2, x2y,x3 + y3), g(f) = (sinf, e')
26. f(x, y) = (x2 - y, y/x, ey), g(s, t, u) = (s + It +
3m, stu)
27. f(x, y, z) = (x + y + z, x3 - evz),
g(s, t, u) = (st, tu, su)
28. Let g: R3 — >• R2 be a differentiable function
such that g(l,-l,3) = (2, 5) and Dg(l,-1,3) =
1 -1 0
0 7
Suppose that f: R — > R is de-
fined by f(x, v) = (2xy, 3x — y + 5). What is
D(fog)(l,-l,3)?
29. Let g:R2^R2 and f:R2^R2 be differentiable
functions such that g(0, 0) = (1, 2), g(l,2) =
(3,5), f(0, 0) = (3, 5), f(4, 1) = (1,2), Z)g(0,0) =
1 0
-1 4
1 1
3 5
Dg{\,2):
Df(4, 1) =
2 3
5 7
£>f(3, 5) =
1 2
1 3
(a) Calculate D(fog)( 1,2).
(b) Calculate D(gof)(4, 1).
30. Let z = f(x,y), where / has continuous partial
derivatives. If we make the standard polar/rectangular
substitution x = r cos 0, y = r sin 9, show that
+
+
1
.30.
2.5 I Exercises 157
31. (a) Use the methods of Example 6 and formula (10)
in this section to determine d2/dx2 and d2/dy2
in terms of the polar partial differential operators
d2/dr2, d2/dd2, d2/dr d6, d/dr, and d/dd. (Hint:
You will need to use the product rule.)
(b) Use part (a) to show that the Laplacian operator
d2/dx2 + d2/dy2 is given in polar coordinates by
the formula
il il
Sx2 + dy2
dr2
+
1 3
r dr
+
l
3<92'
32. Show that the Laplacian operator d2/dx2 + d2/dy2 +
d2/dz2 in three dimensions is given in cylindrical co-
ordinates by the formula
32 d2 d2
dx2 By2 dz2
d2 13 1
■dr-
r dr
+
r2 d62 dz
33. In this problem, you will determine the formula for the
Laplacian operator in spherical coordinates.
(a) First, note that the cylindrical/spherical conver-
sions given by formula (6) of §1.7 express the
cylindrical coordinates z and r in terms of the
spherical coordinates p and <p by equations of pre-
cisely the same form as those that express x and
y in terms of the polar coordinates r and 9. Use
this fact to write 3/3r in terms of 3/3p and d/d<p.
(Also see formula (10) of this section.)
(b) Use the ideas and result of part (a) to establish the
following formula:
32 32 32
d^2 + dy~2 + dz2
+
1 32
3p2 p2 dip2
+
pz sin" cp
so2
2 3 cot go
p dp p2
dcp
34. Suppose that y is defined implicitly as a function y(x)
by an equation of the form
F(x, y) = 0.
(For example, the equation x3 — y2 = 0 defines y as
two functions of x, namely, y = x3/2 and y = — x3/2.
The equation sin(xy) — x2y1 + ey = 0, on the other
hand, cannot readily be solved for y in terms of x. See
the end of §2.6 for more about implicit functions.)
(a) Show that if F and y(x) are both assumed to be
differentiable functions, then
dy_ _ Fx(x, y)
dx Fy(x, y)
provided Fy(x, y) # 0.
(b) Use the result of part (a) to find dy /dx when y
is defined implicitly in terms of x by the equa-
tion x3 — y2 = 0. Check your result by explicitly
solving for y and differentiating.
35. Find dy /dx when y is defined implicitly by the equa-
tion sin(xy) — x2y7 + ey = 0. (See Exercise 34.)
36. Suppose that you are given an equation of the form
F(x,y,z) = Q,
for example, something like x3z + y cosz +
(sin y)/z = 0. Then we may consider z to be defined
implicitly as a function z(x, y).
(a) Use the chain rule to show that if F and z(x , y) are
both assumed to be differentiable, then
3z
dx
Fx(x, y, z)
Fz(x,y,z)'
dz
dy
Fy(x,y,z)
Fz(x, y, z)
(b) Use part (a) to find 3z/3x and 3z/3y where z is
given by the equation xyz = 2. Check your result
by explicitly solving for z and then calculating the
partial derivatives.
37. Find dz/dx and 3z/3y, where z is given implicitly by
the equation
, sin y
x z + y cos z H = 0.
(See Exercise 36.)
38. Let
f(x,y)
2
x y
x2 + y2
0
if (x, y) # (0, 0)
if(x,y) = (0,0)
(a) Use the definition of the partial derivative to find
fx(0, 0) and fy(0, 0).
(b) Let a be a nonzero constant and let x(f) =
(t, at). Show that / o x is differentiable, and find
D(f o x)(0) directly.
(c) Calculate Df(0, O)Z)x(O). How can you reconcile
your answer with your answer in part (b) and the
chain rule?
Let w = f(x, y, z) be a differentiable function of x, y, and
Z. For example, suppose that w = x + 2y + z. Regarding the
variables x, y, and z as independent, we have dw/dx = 1 and
dw/dy = 2. But now suppose that z = xy. Then x, y, and z
are not all independent and, by substitution, we have that w =
X + 2y + xy so that dw/dx = 1 + y and dw/dy = 2 + x. To
overcome the apparent ambiguity in the notation for partial
derivatives, it is customary to indicate the complete set of in-
dependent variables by writing additional subscripts beside
Chapter 2 | Differentiation in Several Variables
the partial derivative. Thus,
/9uA
would signify the partial derivative of w with respect to x,
while holding both y and z constant. Hence, x, y, and z are the
complete set of independent variables in this case. On the other-
hand, we would use (dw/dx)y to indicate thatx andy alone are
the independent variables. In the case that w = x + 2y + z,
this notation gives
9 w
dx
( dw \ ( dw
1, I I = 2, and I
V dy Jx,z V dz
1.
If Z = xy, then we also have
dw\
= 1 + V, and
dx /„
dw
97
2 + x.
In this way, the ambiguity of notation can be avoided. Use this
notation in Exercises 39-45.
39. Let w = x + ly — lOz and z = x2 + y2.
dw\ / dw\ / dw\ f dw
97j,,,; lay A*' \9z)x,y' la*
' dw
(a) Find
and
dy
(b) Relate (dw/dx)yz and (dw/dx)y by using the
chain rule.
40. Repeat Exercise 39 where w = x3 + y2 + z3 and z =
2x — 3y.
41. Suppose s = x2y + xzw — z2 and xyw — v3z + xz
= 0. Find
and
ds
42. Let U = F(P ', V, T) denote the internal energy of a
gas. Suppose the gas obeys the ideal gas law P V = kT,
where & is a constant.
,'dU"
(a) Find
(b) Find
(c) Find
3T
dU
~df
dU
~d~P
43 . Show that if x , y , z are related implicitly by an equation
of the form F(x, y, z) = 0, then
a*\ (dy\ (dz
dy)z \dz)x \dxjv
■1.
This relation is used in thermodynamics. (Hint: Use
Exercise 36.)
44. The ideal gas law PV = kT, where £ is a constant,
relates the pressure P, temperature T, and volume V
of a gas. Verify the result of Exercise 43 for the ideal
gas law equation.
45. Verify the result of Exercise 43 for the ellipsoid
ax2 + by2 + cz2 = d
where a, b, c, and d are constants.
2.6 Directional Derivatives and the Gradient
In this section, we will consider some of the key geometric properties of the
gradient vector
v/ = (i,i *L)
\dx\ dx2 dx„J
of a scalar-valued function of n variables. In what follows, n will usually be 2
or 3.
The Directional Derivative
Let f(x , y) be a scalar- valued function of two variables. In §2.3, we understood the
partial derivative j^{a,b) as the slope, at the point (a, b, f{a, b)), of the curve
obtained as the intersection of the surface z = f(x, y) with the plane y = b.
The other partial derivative ^-(a,b) has a similar geometric interpretation. How-
ever, the surface z = f(x, y) contains infinitely many curves passing through
(a, b, f(a, b)) whose slope we might choose to measure. The directional deriva-
tive enables us to do this.
2.6 | Directional Derivatives and the Gradient
An alternative way to view ^(a,b) is as the rate of change of / as we move
"infmitesimally" from a = (a, b) in the i-direction, as suggested by Figure 2.67.
This is easy to see since, by the definition of the partial derivative,
df< « r f(a + h,b)-f(a,b)
-(a, b) = hm
dx h^O
lim
lim
f((a,b) + (h,0))- f(a,b)
f((a,b) + h(l,0))- f(a,b)
o h
r /(a + M)-/(a)
= lim .
h-+Q h
Note that we are identifying the point (a, b) with the vector a = {a, b) = a\ + bj.
Similarly, we have
V, .. /(a + /ij)-/(a)
— (a, b) = lim .
dy h^o h
Writing partial derivatives as we just have enables us to see that they are
special cases of a more general type of derivative. Suppose v is any unit vector in
R2. (The reason for taking a unit vector will be made clear later.) The quantity
lim/(a + <,v)-/(a)
h^O h
is nothing more than the rate of change of / as we move (infmitesimally) from a =
(a , b) in the direction specified by v = (A, B) = Ai + By It's also the slope of the
curve obtained as the intersection of the surface z = f{x,y) with the vertical plane
B(x — a) — A(y — b) = 0. (See Figure 2.68.) We can use the limit expression in
(1) to define the derivative of any scalar- valued function in a particular direction.
Figure 2.67 Another way to view the
partial derivative 9 // dx at a point.
Figure 2.68 The directional derivative.
Chapter 2 | Differentiation in Several Variables
DEFINITION 6.1 Let X be open in R", /:XCR"->Ra scalar-valued
function, and a e X . If v e R" is any unit vector, then the directional deriva-
tive of / at a in the direction of v, denoted Dv/(a), is
/(a + /*v)-/(a)
Dv/(a) = hm
h^O h
(provided that this limit exists).
EXAMPLE 1 Suppose f(x, y) = x2 - 3xy + 2x - 5y. Then, if v = (v, w) e
R2 is any unit vector, it follows that
f((0,0) + h(v,w))-f(0, 0)
Dyf(0, 0) = lim
h^O h
2v2 — 3h2vw + 2hv — 5hw
= lim
h^o h
= lim(7ii;2 — 3h vw + 2v — 5w)
= 2v — 5w.
Thus, the rate of change of / is 2v — 5w if we move from the origin in the
direction given by v. The rate of change is zero if v = (5/V29, 2/V29) or
(-5/V29, -2/V59). ♦
Consequently, we see that the partial derivatives of a function are just the "tip
of the iceberg." However, it turns out that when / is differentiable, the partial
derivatives actually determine the directional derivatives for all directions v. To
see this rather remarkable result, we begin by defining a new function F of a
single variable by
F(0 = /(a-Mv).
Then, by Definition 6. 1, we have
n f( . Y /(a + rv)-/(a) F(t) - F(0)
Dv/(a) = hm = hm — = F (0).
t^O t r^O t — 0
That is,
Ov/(a) = ^/(a + *v)|f=0. (2)
The significance of equation (2) is that, when / is differentiable at a, we can apply
the chain rule to the right-hand side. Indeed, let x(r) = a + fv. Then, by the chain
rule,
^/(a + fv) = Df(x)Dx(t) = D/(x)v.
dt
Evaluation at t = 0 gives
Dv/(a) = Z)/(a)v = V/(a).v. (3)
The purpose of equation (3) is to emphasize the geometry of the situation. The
result above says that the directional derivative is just the dot product of the
2.6 | Directional Derivatives and the Gradient
gradient and the direction vector v. Since the gradient is made up of the partial
derivatives, we see that the more general notion of the directional derivative
depends entirely on just the direction vector and the partial derivatives. To be
more formal, we summarize this discussion with a theorem.
THEOREM 6.2 Let X C R" be open and suppose /: X -> R is differentiable
at a e X. Then the directional derivative Dv/(a) exists for all directions (unit
vectors) veR" and moreover, we have
Dv/(a) = V/(a).v.
EXAMPLE 2 The function f(x, y) = x2 — 3xy + 2x — 5y we considered in
Example 1 has continuous partials and hence, by Theorem 3.5, is differentiable.
Thus, Theorem 6.2 applies to tell us that, for any unit vector v = i>i + w\ 6 R2,
Dv/(0, 0) = V/(0, 0) • v = (fx(0, 0)i + /,(0, 0)j) • (vi + wj)
= (2i-5j)-(ui+Wj)
= 2v — 5w,
as seen earlier. ♦
EXAMPLE 3 The converse of Theorem 6.2 does not hold. That is, a function
may have directional derivatives in all directions at a point yet fail to be differ-
entiable. To see how this can happen, consider the function /: R2 -> R denned
by
/(*. y) =
if (jc, y) * (0, 0)
xL + y4
0 if(je,y) = (0,0)
This function is not continuous at the origin. (Why?) So, by Theorem 3.6, it
fails to be differentiable there; however, we claim that all directional derivatives
exist at the origin. To see this, let the direction vector v be vi + wj. Hence, by
Definition 6.1, we observe that
Dv/(0, 0) = lim
f((0,0) + Kvi + wj))-f(0, 0)
h
= lim
h^0
hv(hw)
{hvf + (hw)4
h2vw2
lim
h^o h2(v2 + h2w4)
,2 „,„2
= lim
vw
h^0 V2 + h2w4
vw
w
V
Chapter 2 | Differentiation in Several Variables
Thus, the directional derivative exists whenever v 7^ 0. When v = 0 (in which
case v = j), we, again, must calculate
/((0, 0) + fcj)-/(0, 0)
Dj/(0, 0)=lim
h-*o h
= hm
fc->-0
0-0
= hm = 0.
ft->-o h
Consequently, this directional derivative (which is, in fact, df/dy) exists as well.
♦
The reason we have restricted the direction vector v to be of unit length in our
discussion of directional derivatives has to do with the meaning of Dv/(a), not
with any technicalities pertaining to Definition 6.1 or Theorem 6.2. Indeed, we
can certainly define the limit in Definition 6. 1 for any vector v, not just one of unit
length. So, suppose w is an arbitrary nonzero vector in R" and / is differentiable.
Then the proof of Theorem 6.2 goes through without change to give
hm = V / (a) • w.
h-*o h
The problem is as follows: If w = kv for some (nonzero) scalar k, then
hm = V / (a) • w
h-f0 h
= V/(a)-(*v)
= *(V/(a)-v)
n /(a + Av)-/(a)
That is, the "generalized directional derivative" in the direction of kv is k times
the derivative in the direction of v. But v and kv are parallel vectors, and it is
undesirable to have this sort of ambiguity of terminology. So we avoid the trouble
by insisting upon using unit vectors only (i.e., by allowing k to be ±1 only) when
working with directional derivatives.
Gradients and Steepest Ascent
Suppose you are traveling in space near the planet Nilrebo and that one of your
spaceship's instruments measures the external atmospheric pressure on your ship
as a function f(x, y, z) of position. Assume, quite reasonably, that this function
is differentiable. Then Theorem 6.2 applies and tells us that if you travel from
point a = (a , b, c) in the direction of the (unit) vector u = ui + uj + 10k, the rate
of change of pressure is given by
Du/(a) = V/(a).u.
Now, we ask the following: In what direction is the pressure increasing the most?
If 9 is the angle between u and the gradient vector V/(a), then we have, by
Theorem 3.3 of §1.3, that
Dn/(a) = ||V/(a)|| ||u|| costf = ||V/(a)|| cos#,
2.6 | Directional Derivatives and the Gradient 163
since u is a unit vector. Because — 1 < cos 6 < 1 , we have
-||V/(a)|| < Du/(a)< ||V/(a)||.
Moreover, cos 9 = 1 when 9 = 0 and cos 9 = — 1 when 9 = it . Thus, we have
established the following:
THEOREM 6.3 The directional derivative D„f(a) is maximized, with respect to
direction, when u points in the same direction as V/(a) and is minimized when u
points in the opposite direction. Furthermore, the maximum and minimum values
of Du/(a) are || V/(a)|| and -|| V/(a)||, respectively.
EXAMPLE 4 If the pressure function on Nilrebo is
f(x, y, z) = 5x2 + 7y4 + x2z2 arm,
where the origin is located at the center of Nilrebo and distance units are measured
in thousands of kilometers, then the rate of change of pressure at (1,-1,2)
in the direction of i + j + k may be calculated as V/(l, — 1, 2) • u, where u =
(i + j + k)/\/3. (Note that we normalized the vector i + j + k to obtain a unit
vector.) Using Theorem 6.2, we compute
A./(l,-l,2) = V/(l,-l,2).u
i + j + k
= (18i - 28j + 4k) V—
18-28 + 4
V3
2\/3 atm/Mm.
Additionally, in view of Theorem 6.3, the pressure will increase most rapidly
in the direction of V/(l , —1,2), that is, in the
18i-28j+4k 9i-14j + 2k
||18i-28j+4k|| V28T
direction. Moreover, the rate of this increase is
|| V/(l, -1,2)|| = 2^/281 atm/Mm. ♦
Theorem 6.3 is stated in a manner that is independent of dimension — that is, so
that it applies to functions /: X C R" —> R for any n > 2. In the case n = 2, there
is another geometric interpretation of Theorem 6.3: Suppose you are mountain
climbing on the surface z = fix, y). Think of the value of / as the height of the
mountain above (or below) sea level. If you are equipped with a map and compass
(which supply information in the xy-plane only), then if you are at the point on
the mountain with xy-coordinates (map coordinates) (a, b), Theorem 6.3 says
that you should move in the direction parallel to the gradient V/(a , b) in order to
climb the mountain most rapidly. (See Figure 2.69.) Similarly, you should move
in the direction parallel to — V f(a , b) in order to descend most rapidly. Moreover,
the slope of your ascent or descent in these cases is || V/(a, b)\\ . Be sure that you
understand that V/(a, b) is a vector in R2 that gives the optimal north-south,
east-west direction of travel.
1 64 Chapter 2 | Differentiation in Several Variables
Tangent Planes Revisited
In §2.1, we indicated that not all surfaces can be described by equations of the
form z = f(x, y). Indeed, a surface as simple and familiar as the sphere is not the
graph of any single function of two variables. Yet the sphere is certainly smooth
enough for us to see intuitively that it must have a tangent plane at every point.
(See Figure 2.70.)
How can we find the equation of the tangent plane? In the case of the unit
sphere x2 + y2 + z2 = 1, we could proceed as follows: First decide whether the
point of tangency is in the top or bottom hemisphere. Then apply equation (4) of
Figure 2.70 A sphere and one of §2.3 to the graph of z = — x2 — y2 or z = —y/\ — x2 — y2, as appropriate.
its tangent planes. The calculus is tedious but not conceptually difficult. However, the tangent planes
to points on the equator are all vertical and so equation (4) of §2.3 does not apply.
(It is possible to modify this approach to accommodate such points, but we will
not do so.) In general, given a surface described by an equation of the form
F(x, v, z) = c (where c is a constant), it may be entirely impractical to solve
for z even as several functions of x and y. Try solving for z in the equation
xyz + ye" — x2 + yz2 = 0 and you'll see what we mean. We need some other
way to get our hands on tangent planes to surfaces described as level sets of
functions of three variables.
To get started on our quest, we present the following result, interesting in its
own right:
THEOREM 6.4 Let X C R" be open and /: X -> R be a function of class C1.
If x0 is a point on the level set S = {x e X \ f(x) = c}, then the vector V/(x0)
is perpendicular to S.
2.6 | Directional Derivatives and the Gradient
PROOF We need to establish the following: If v is any vector tangent to S at xo,
then V/(xo) is perpendicular to v (i.e., V/(xo) • v = 0). By a tangent vector to S
at xo, we mean that v is the velocity vector of a curve C that lies in S and passes
through Xq. The situation in R3 is pictured in Figure 2.71.
Figure 2.71 The level set surface
S = {x | /(x) = c}.
Thus, let C be given parametrically by x(t) = (xi(/), X2(t), . . . , x„(t)), where
a < t < b and x(fo) = xo for some number to in (a, b). (Then, if v is the velocity
vector at x0, we must have x'(f0) = v. See §3.1 for more about velocity vectors.)
Since C is contained in S, we have
/(x(0) = /(*i(0. *2(0. • ■ • . *»(0) = c-
Hence,
jt imm = jfc\ = o. (4)
On the other hand the chain rule applied to the composite function fox:
(a,b) -> R tells us
^[/(x(f))] = V/(x(0)-x'(0-
Evaluation at to and equation (4) let us conclude that
V/(x(/0))-x'(*o) = V/(x0)-v = 0,
as desired. ■
Here's how we can use the result of Theorem 6.4 to find the plane tan-
gent to the sphere x2 + y2 + z2 = 1 at the point ^— 0, ^J. From §1.5, we
know that a plane is determined uniquely from two pieces of information: (i) a
point in the plane and (ii) a vector perpendicular to the plane. We are given a
point in the plane in the form of the point of tangency ^— -7^, 0, ^=V As for
a vector normal to the plane, Theorem 6.4 tells us that the gradient of the func-
tion f(x, y, z) = x2 + y2 + z2 that defines the sphere as a level set will do. We
have
V f(x, y, z) = 2xi + 2y\ + 2zk,
so that
v/(-7!'°'7!) = ^'+V5k-
Chapter 2 | Differentiation in Several Variables
Hence, the equation of the tangent plane is
V2f, + -L) + ^--L)=0,
= 0,
or
X = y/2.
In general, if S is a surface in R3 defined by an equation of the form
f(x,y,z) = c,
then if xo £ X, the gradient vector V/(xo) is perpendicular to S and, conse-
quently, if nonzero, is a vector normal to the plane tangent to S at x0. Thus,
the equation
V/(x0)-(x-x0) = 0
or, equivalently,
fx(x0, y0, z0)(x - x0) + fy(x0, y0, zo)(y - v0)
+ fz(*o, yo, zo)(z - zo) = 0
is an equation for the tangent plane to S at x0.
(5)
(6)
Note that formula (5) can be used in R" as well as in R3, in which case it
defines the tangent hyperplane to the hypersurface S C R" defined by f{x\ ,
X2, ■ ■ ■ , xn) = c at the point xo e S.
EXAMPLE 5 Considerthe surface S defined by the equation*3 y — yz2 + z5 =
9. We calculate the plane tangent to S at the point (3,-1,2).
To do this, we define f(x, y, z) = x3y — yz2 + z5- Then
V/(3, -1,2)= (3x2yi + (x3 - z2)\ + (5Z4 - 2yz)k)\(3 _h2)
= -21\ + 23j + 84k
is normal to S at (3, —1, 2) by Theorem 6.4. Using formula (6), we see that the
tangent plane has equation
or, equivalently,
-270 - 3) + 23(y + 1) + 84(z - 2) = 0
-27x + 23y + 84z = 64.
EXAMPLE 6 Consider the surface defined by z4 = x2 + y2. This surface is
the level set (at height 0) of the function
f{x, y, Z) = x2 + y2- z4.
The gradient of / is
Vf(x, y,z) = 2xi + 2yj- 4z3 k.
2.6 | Directional Derivatives and the Gradient
Figure 2.72 The
surface of Example 6.
Note that the point (0, 0, 0) lies on the surface. However, V/(0, 0, 0) = 0, which
makes the gradient vector unusable as a normal vector to a tangent plane. Thus,
formula (6) doesn't apply. What we conclude from this example is that the surface
fails to have a tangent plane at the origin, a fact that is easy to believe from the
graph. (See Figure 2.72.) ♦
EXAMPLE 7 The equation x2 + y2 + z2 + w2 = 4 defines a hypersphere of
radius 2 in R4. We use formula (5) to determine the hyperplane tangent to the
hypersphere at (— 1 , 1 , 1,-1).
The hypersphere may be considered to be the level set at height 4 of the
function f(x, y, z, w) = x2 + y2 + z2 + w2, so that the gradient vector is
V f(x, y, z, w) = (2x, 2y, 2z, 2w),
so that
V/(-l, 1,1,-1) = (-2, 2, 2, -2).
Using formula (5), we obtain an equation for the tangent hyperplane as
(-2,2,2, -2)-(x + l,y- 1, z - 1, w + 1) = 0
or
-2(x + 1) + 2(y - 1) + 2(z - 1) - 2(w + 1) = 0.
Equivalently, we have the equation
x - y - z + w + 4 = 0. ♦
EXAMPLE 8 We determine the plane tangent to the paraboloid z
- v2
3y
at the point (—2, 1, 7) in two ways: (i) by using formula (4) in §2.3, and (ii) by
using our new formula (6).
First, the equation z = x2 + 3y2 explicitly describes the paraboloid as the
graph of the function f(x, y) = x2 + 3y2, that is, by an equation of the form
z = fix, y). Therefore, formula (4) of §2.3 applies to tell us that the tangent
plane at (—2, 1, 7) has equation
z = f(-2, 1) + fx(-2, l)(x + 2) + fy(-2, l)(y - 1)
or, equivalently,
z = 7 - 4(jc + 2) + 6(y - 1). (7)
Second, if we write the equation of the paraboloid as x2 + 3y2 — z = 0,
then we see that it describes the paraboloid as the level set of height 0 of the
three-variable function F(x, y, z) = x2 + 3y2 — z. Hence, formula (6) applies
and indicates that an equation for the tangent plane at (—2, 1, 7) is
Fx(-2, 1, 7)(jc + 2) + Fy{-2, 1, 7)(y - 1) + Fz(-2, 1, 7)(z - 7) = 0
or
-4(x + 2) + 6(y- 1) - l(z - 7)
As can be seen, equation (7) agrees with equation (8).
(8)
♦
Example 8 may be viewed in a more general context. If S is the surface in R3
given by the equation z = f(x, y) (where / is differentiate), then formula (4) of
§ 2 . 3 tells us that an equation for the plane tangent to S at the point (a , b , / {a , b)) is
z = f(a, b) + fx(a, b)(x -a)+ Ua, b)(y - b).
Chapter 2 | Differentiation in Several Variables
(-2,2,6)
Figure 2.73 The two-sheeted
hyperboloid z2/4 — x2 — y2 = 1.
The point (—2, 2, 6) lies on the
sheet given by z = 2^/x2 + y2 + 1,
and the point (1,1, — 2^/3) lies on
the sheet given by
z = -2jx2 + y2+l.
1.
At the same time, the equation for S may be written as
f(x, y) - z = 0.
Then, if we let F(x, y, z) = f(x, y) — z, we see that S is the level set of F at
height 0. Hence, formula (6) tells us that the tangent plane at (a, b, f(a, b)) is
Fx(a, b, f(a, b))(x - a) + Fy(a, b, f(a, b))(y - b)
+ Fe(a, b, f(a, b))(z - f(a, b)) = 0.
By construction of F,
dF _df dF _ df dF
dx dx ' dy dy ' dz
Thus, the tangent plane formula becomes
fx(a, b)(x -a) + fy(a, b)(y - b) - (z - f(a, b)) = 0.
The last equation for the tangent plane is the same as the one given above by
equation (4) of §2.3.
The result shows that equations (5) and (6) extend the formula (4) of §2.3 to
the more general setting of level sets.
The Implicit Function and Inverse Function
Theorems (optional)
We have previously noted that not all surfaces that are described by equations of
the form F(x ,y,z) = c can be described by an equation of the form z = f(x , y).
We close this section with a brief — but theoretically important — digression about
when and how the level set {(x, y, z) | F(x, y, z) = c} can also be described as
the graph of a function of two variables, that is, as the graph of z = f(x, y).
We also consider the more general question of when we can solve a system of
equations for some of the variables in terms of the others.
We begin with an example.
EXAMPLE 9 Consider the hyperboloid z2/4 - x2
described as the level set (at height 1) of the function
y = 1, which may be
F{x, y, z) =
y
(See Figure 2.73.) This surface cannot be described as the graph of an equation
of the form z = f(x, y), since particular values for x and y give rise to two values
for z. Indeed, when we solve for z in terms of x and y, we find that there are two
functional solutions:
x 2 + y2 + 1 and z
= -2/
x2 + y2 + 1.
(9)
On the other hand these two solutions show that, given any particular point
(*o, yo, zo) of the hyperboloid, we may solve locally for z in terms of x and y.
That is, we may identify on which sheet of the hyperboloid the point {xq, yo, Zo)
lies and then use the appropriate expression in (9) to describe that sheet. ♦
Example 9 prompts us to pose the following question: Given a surface S,
described as the level set {(x, y, z) | F(x, y, z) = c], can we always determine at
least a portion of S as the graph of a function z = f(x , y)? The result that follows,
2.6 | Directional Derivatives and the Gradient
a special case of what is known as the implicit function theorem, provides
relatively mild hypotheses under which we can.
THEOREM 6.5 (The implicit function theorem) Let F:XcR"->R be
of class C1 and let a be a point of the level set S = {x 6 R" | F(x) = c}. If
FXn(a) 0, then there is a neighborhood U of (a\, ci2, . . . , a„_i) in Rn_1, a
neighborhood V of an in R, and a function f:U<^ R"_1 -> V of class C1
such that if (xi, X2, ■ ■ ■ , xn-\) e U andx„ e V satisfy F(x\, X2, ■ • ■ , xn) = c (i.e.,
(x\,x2, x„) e S), thenx„ = f(xi,x2, x„_i).
The significance of Theorem 6.5 is that it tells us that near a point a € S
such that dF/dxn ^ 0, the level set S given by the equation F{x\, . . . , xn) = c
is locally also the graph of a function x„ = /(xj , . . . , x„_i). In other words, we
may solve locally for xn in terms of x\, . . . , x„_i, so that S is, at least locally, a
differentiable hypersurface in R" .
EXAMPLE 10 Returning to Example 9, we recall that the hyperboloid is the
level set (at height 1) of the function F(x, y, z) = z2/4 — x2 — y2. We have
dF
9z"
Note that for any point (xo, yo> Zo) in the hyperboloid, we have |zol > 2. Hence,
9F.(xo, yo, zo) 0. Thus, Theorem 6.5 implies that we may describe a portion
of the hyperboloid near any point as the graph of a function of two variables. This
is consistent with what we observed in Example 9. ♦
Of course, there is nothing special about solving for the particular vari-
able x„ in terms of Suppose a is a point on the level set S de-
termined by the equation F(x) = c and suppose VF(a) / 0. Then FXj(a) 0 for
some i . Hence, we can solve locally near a for x,- as a differentiable function of
x\, . . . , x,_i, Xi+i, . . . , x„. Therefore, S is locally a differentiable hypersurface
inR".
EXAMPLE 11 Let S denote the ellipsoid x2/4 + y2/36 + z2/9
is the level set (at height 1) of the function
1. Then S
n*,y,z)=^ + y- +
At the point (V2, V6, V3), we have
dF
~dz
2z
(V2.V6.V3)
2^3
(V2,V6,V3)
Thus, S may be realized near (V2, V6, V3) as the graph of an equation of the
form z = f(x, y), namely, z = 3^/1 - x2/4 — y2/36. At the point (0, —6, 0),
however, we see that dF /dz vanishes. On the other hand,
dF
97
(0,-6,0)
2±
36
(0,-6,0)
^0.
Consequently, near (0, —6, 0), the ellipsoid may be described by solving for y as
a function of x and z, namely, y = —6^/ 1 — x2/4 — z2/9. ♦
Chapter 2 | Differentiation in Several Variables
EXAMPLE 1 2 Consider the set of points S defined by the equation x2z2 — y =
0. Then S is the level set at height 0 of the function F(x, y, z) = x2z2 — y. Note
that
VF(x,y,z) = (2xz2, -l,2x2z).
Since dF /dy never vanishes, we see that we can always solve for y as a function
of x and z. (This is, of course, obvious from the equation.) On the other hand, near
points where x and z are nonzero, both dF/dx and dF/dz are nonzero. Hence,
we can solve for either x or z in this case. For example, near (1 , 1 , — 1), we have
x=H and z=-Ii-
As just mentioned, Theorem 6.5 is actually a special case of a more general
result. In Theorem 6.5 we are attempting to solve the equation
F(xi, x2, . . . , xn) = c
for xn in terms of x\, . . . , xn-\. In the general case, we have a system of m
equations
F\(x\, . . . , x„, yi, . . . , ym) = c1
F2(xi,...,x„,yi,...,ym) = c2
^mv^l ' • ■ ■ ' 3^1) ■ ■ ■ ' y?n)
and we desire to solve the system for yi , . . . , ym in terms of x\ , . . . , x„. Using vec-
tor notation, we can also write this system as F(x, y) = c, where x = (xi, . . . , xn),
y = (yi, . . . , ym), c = (ci , . . . , cm), and Fi, ... , Fm make up the component
functions of F. With this notation, the general result is the following:
THEOREM 6.6 (The implicit function theorem, general case) Suppose
F: A ->• Rm is of class C1, where A is open in R"+m. Let (a, b) = (a\, . . . , a„,
b\, . . . , bm) £ A satisfy F(a, b) = c. If the determinant
r 3Fi
dy
A(a, b) = det
(a,b)
9F,
dy,
-(a, b)
dFm
L dy
-(a, b)
dFm
dyr.
(a, b)
^0,
then there is a neighborhood U of a in R" and a unique function f : U —> Rm of
class C1 such that f(a) = b and F(x, f(x)) = c for all x e U. In other words, we
can solve locally for y as a function f(x).
EXAMPLE 13 We show that, near the point (x\, x%, x^, y\, y2) = (— 1, 1, 1,
2, 1), we can solve the system
x\y2 + x2y\ = 1
xfx3yi + x2y{ = 3
for y\ and y2 in terms of x\ , x2, x%.
2.6 | Directional Derivatives and the Gradient
We apply the general implicit function theorem (Theorem 6.6) to the system
IFi(xi, x2, x3, y\,yi) = x\y2 + x2yi = 1
F2(xi,x2, x3, y\,y2) = xfx3yi + x2y\ = 3
The relevant determinant is
A(-l, 1, 1,2, 1) = det
= det
= det
3Fi dFi
dyi dy2
dF2 dF2
dyi dy2
x2 X\
(X1,.T2,X3,3'1,}'2)=(- 1,1, 1,2,1)
xfxj, 3x2yj
(xi ,a-2, x3 , vi , y2)={- 1 , 1 , 1 ,2, 1)
4/0.
Hence, we may solve locally, at least in principle.
We can also use the equations in (11) to determine, for example,
dy2
(— 1, 1, 1), where we treat x\, x2, x3 as independent variables and y\ and
dx\
y2 as functions of them.
Differentiating the equations in (1 1) implicitly with respect to X\ and using
the chain rule, we obtain
yi + xx- — Vx2—
ax\ ax\
0
Ixix^yi +x\xt,- r-3x2y;
dy2
= 0
.2„
Now, let (xi , x2, Xi, y\ , y2) — (— 1, 1 , 1,2, 1), so that the system becomes
f*(-l,l,l)-|^(-l,l,l) = -l
3xi 6x\
^l(-l,l,l) + 3^(-l,l,l) = 4
3xi dxi
3y2 5
We may easily solve this last system to find that — (—1,1,1) = —. ♦
3xi 4
Now, suppose we have a system of n equations that defines the variables
yi , . . . , yn in terms of the variables xi, . . . , x„, that is,
yi = /i(xi, . . . ,x„)
yi = fi(xu ■ . . ,xn)
y„ = fn(X\,...,Xn)
(12)
Note that the system given in (12) can be written in vector form as y = f(x). The
question we ask is, when can we invert this system? In other words, when can we
Chapter 2 | Differentiation in Several Variables
solve for x\ , . . . ,xn in terms of yi ,
function g so that x = g(y)?
The solution is to apply Theorem 6.6 to the system
yn, or, equivalently, when can we find a
F2(xi,
,xn,y\,
x„,y{,
.xn,y\,
y„) = 0
y„) = o
Jn) = 0
where Ff(jti, . . . , xn, y\, . . . , y„) = fi(xi, . . . , xn) — y,-. (In vector form, we are
setting F(x, y) = f(x) — y.) Then solvability for x in terms of y near x = a, y = b
is governed by the nonvanishing of the determinant
detDf(a) = det
This determinant is also denoted by
3(/i-
3xi
dxt
(a)
(a)
dxn
dfn
dxn
(a)
(a)
d(xu . . . ,x„)
and is called the Jacobian of f = (/i, . . . , /„). A more precise and complete
statement of what we are observing is the following:
THEOREM 6.7 (The inverse function theorem) Suppose f = (f\ /„)
is of class C1 on an open set A c R" . If
then there is an open set U c R" containing a such that f is one-one on U, the
set V = f(U) is also open, and there is a uniquely determined inverse function
g: V -> U to f, which is also of class C1 . In other words, the system of equations
y = f(x) may be solved uniquely as x = g(y) for x near a and y near b.
det Df(a)
3(/i,..../»)
3(xi, . . . , xn)
EXAMPLE 14 Consider the equations that relate polar and Cartesian coordi-
nates:
' x = r cos#
y = r sin 6
These equations define x and y as functions of r and 6. We use Theorem 6.7 to
see near which points of the plane we can invert these equations, that is, solve for
r and 9 in terms of x and y.
To use Theorem 6.7, we compute the Jacobian
d(x, y)
d(r, 9)
cos 6 —rsmO
sin 0 r cos 9
r.
Thus, we see that, away from the origin (r = 0), we can solve (locally) for r and 9
uniquely in terms of x and y. At the origin, however, the inverse function theorem
2.6 I Exercises 173
does not apply. Geometrically, this makes perfect sense, since at the origin the
polar angle 9 can have any value. ♦
2.6 Exercises
1 . Suppose f(x , y , z) is a differentiable function of three
variables.
(a) Explain what the quantity V/(x, y, z) • (— k) rep-
resents.
(b) How does V/(x, y, z) • (-k) relate to 3//3z?
In Exercises 2—8, calculate the directional derivative of the
given function f at the point a in the direction parallel to the
vector u.
2. f(x, y) = ev sinx, a = o), u
31- j
10
3. f(x, y) = x2- 2x3y + 2y\ a = (2, -1), u =
4. f{x, y) =
1
-,a = (3, -2),u = i-j
(x2 + y2)'
5. /(x, y) = ex - x2y, a = (1, 2), u = 2i + j
6. f(x, y, z) = xyz, a = (-1, 0, 2), u
2k- i
~7f
7. /(x, y, z) = e~
8- f(x,y,z)
3k
9. For the function
/(*. y)
3z2 + l
+z \a = (l,2,3),u = i + j + k
, a = (2, -1,0), u = i-2j +
x\y\
0
if (x, y) ? (0, 0)
if(x,y) = (0, 0)
(a) calculate fx(0, 0) and fy(0, 0). (You will need to
use the definition of the partial derivative.)
(b) use Definition 6.1 to determine for which unit
vectors v = ui + urj the directional derivative
Dv/(0, 0) exists.
(c) use a computer to graph the surface z = f(x, y).
1 0. For the function
XV
Jx2 + .
if(x,y)/(0, 0)
if(x,y) = (0, 0)
(a) calculate /v(0, 0) and /,(0, 0).
(b) use Definition 6.1 to determine for which unit
vectors v = vi + w\ the directional derivative
Dv/(0, 0) exists.
(c) use a computer to graph the surface z = f(x, y).
1 1 . The surface of Lake Erehwon can be represented by a
region D in the jcv-plane such that the lake's depth (in
meters) at the point (x, y) is given by the expression
400 — 3x2y2. If your calculus instructor is in the wa-
ter at the point (1, —2), in which direction should she
swim
(a) so that the depth increases most rapidly (i.e., so
that she is most likely to drown)?
(b) so that the depth remains constant?
12. A ladybug (who is very sensitive to temperature) is
crawling on graph paper. She is at the point (3,7) and
notices that if she moves in the i-direction, the tem-
perature increases at a rate of 3 deg/cm. If she moves
in the j -direction, she finds that her temperature de-
creases at a rate of 2 deg/cm. In what direction should
the ladybug move if
(a) she wants to warm up most rapidly?
(b) she wants to cool off most rapidly?
(c) she desires her temperature not to change?
13. You are atop Mt. Gradient, 5000 ft above sea level,
equipped with the topographic map shown in Fig-
ure 2.74. A storm suddenly begins to blow, necessitat-
ing your immediate return home. If you begin heading
due east from the top of the mountain, sketch the path
that will take you down to sea level most rapidly.
14. It is raining and rainwater is running off an ellipsoidal
dome with equation Ax2 + y2 + 4z2 = 16, where
z > 0. Given that gravity will cause the raindrops to
slide down the dome as rapidly as possible, describe
the curves whose paths the raindrops must follow.
(Hint: You will need to solve a simple differential
equation.)
15. Igor, the inchworm, is crawling along graph paper in
a magnetic field. The intensity of the field at the point
(x, y) is given by M(x, y) = 3x2 + y2 + 5000. If Igor
is at the point (8, 6), describe the curve along which he
should travel if he wishes to reduce the field intensity
as rapidly as possible.
In Exercises 16— 19, find an equation for the tangent plane to
the surface given by the equation at the indicated point (xo , yo ,
zo)-
1 74 Chapter 2 | Differentiation in Several Variables
Figure 2.74 The topographic map of Mt. Gradient in Exercise 13.
16. x1 + y3 + ;3 = 7, (xq, y0, zo) = (0, -1, 2)
17. zey cosx = 1, (x0, yo, zo) = (x, 0, -1)
18. 2xz + yz- x2y + 10 = 0, Oo, yo, zo) = (1, -5, 5)
19. 2xy2 = 2z2 - xyz, (x0, y0, Zo) = (2, -3, 3)
20. Calculate the plane tangent to the surface whose equa-
tion is x2 — 2y2 + 5xz = 7 at the point (—1, 0, — |)in
two ways:
(a) by solving for z in terms of x and y and using
formula (4) in §2.3
(b) by using formula (6) in this section.
21. Calculate the plane tangent to the surface xsiny +
xz2 = 2eyz at the point (2, j, 0) in two ways:
(a) by solving for x in terms of y and z and using a
variant of formula (4) in §2.3
(b) by using formula (6) in this section.
22. Find the point on the surface x3 — 2y2 + z2 = 21
where the tangent plane is perpendicular to the line
given parametrically as x = 3f — 5, v = 2t + 7, z =
1 - V2f.
23. Find the points on the hyperboloid 9x2 — 45y2 +
5z2 = 45 where the tangent plane is parallel to the
plane x + 5y — 2z = 7.
24. Show that the surfaces z = lx2 — \2x — 5y2 and
xyz
2 intersect orthogonally at the point (2, 1 , — 1).
25. Suppose that two surfaces are given by the equations
Moreover, suppose that these surfaces intersect at the
point (jco, yo, Zo)- Show that the surfaces are tangent at
(x0, yo, Zo) if and only if
VF(x0, yo, zo) x VG(i0, y0, zo) = 0.
26. Let S denote the cone x2 + Ay2 = z2 ■
(a) Find an equation for the plane tangent to S at the
point (3, —2, —5).
(b) What happens if you try to find an equation for a
tangent plane to S at the origin? Discuss how your
findings relate to the appearance of S.
27. Consider the surface S defined by the equation x3 —
x2y2 + Z2 = 0.
(a) Find an equation for the plane tangent to S at the
point (2, -3/2, 1).
(b) Does S have a tangent plane at the origin? Why or
why not?
If a curve is given by an equation of the form fix, y) = 0, then
the tangent line to the curve at a given point (xo, yo) °n it niay
be found in two ways: (a) by using the technique of implicit
differentiation from single-variable calculus and (b) by using
a formula analogous to formula (6). In Exercises 28-30, use
both of these methods to find the lines tangent to the given
curves at the indicated points.
28. x2 + y2 = 4, (xo, y0) = (-V2, V2)
29. y3 =x2+x\(x0,yo) = (1,^/2)
F(x, y, z) = c and G(x, y, z) = k.
30. x5 + 2xy + y3 = 16, (x0, y0) = (2, -2)
2.6 I Exercises 175
Let C be a curve in R2 given by an equation of the form
f(x, y) = 0. The normal line to C at a point (xo, yo) on it
is the line that passes through (xo, yo) and is perpendicular
to C (meaning that it is perpendicular to the tangent line to
C at (xo, yo)). In Exercises 31-33, find the normal lines to
the given curves at the indicated points. Give both a set of
parametric equations for the lines and an equation in the form
Ax + By = C. (Hint: Use gradients.)
31. x2-y2 = 9,(x0,y0) = (5, -4)
32. x2 - x3 = y2, (x0, yo) = (-1, 72)
33. x3 - 2xy + y5 = 11, (x0, y0) = (2, -1)
34. This problem concerns the surface defined by the equa-
tion
x3z + x2y2 + sin(yz) = -3.
(a) Find an equation for the plane tangent to this sur-
face at the point (—1, 0, 3).
(b) The normal line to a surface S in R3 at a point
(xo, yo, Zo) on it is the line that passes through
(xo, yo, Zo) and is perpendicular to S. Find a set
of parametric equations for the line normal to the
surface given above at the point (—1,0, 3).
35. Give a set of parametric equations for the normal line to
the surface defined by the equation exy + exz — 2eyz =
0 at the point (-1,-1,-1). (See Exercise 34.)
36. Give a general formula for parametric equations for
the normal line to a surface given by the equation
F(x, y, z) = 0 at the point (xo, yo, zo) on the surface.
(See Exercise 34.)
37. Generalizing upon the techniques of this section,
find an equation for the hyperplane tangent to
the hypersurface sinxi + cosx2 + sinx3 + cos X4 +
sinxs = — 1 at the point (jr, jt, 3jt/2, 2jt, 2jt) e R5.
38. Find an equation for the hyperplane tangent to the
(« — l)-dimensional ellipsoid
2,0 2,0 2 , , 2 »(" + 1)
X[ + 2x2 + 3x3 H h nxn =
at the point (-1, -1,..., -1) e R".
39. Find an equation for the tangent hyperplane to the (n —
l)-dimensional sphere x\ + x\ + ■ ■ ■ + x2 = 1 in R"
at the point (1 j+Jn, \/-Jn, . . . , l/Vn, — l/*/n).
Exercises 40—49 concern the implicit function theorems and
the inverse function theorem (Theorems 6.5, 6.6, and 6. 7).
40. Let S be described by z2y3 + x2y = 2.
(a) Use the implicit function theorem to determine
near which points S can be described locally as
the graph of a C1 function z = /(x, y).
(b) Near which points can S be described (locally) as
the graph of a function x = g(y, z)?
(c) Near which points can S be described (locally) as
the graph of a function y = h(x, z)?
41 . Let S be the set of points described by the equation
sinxy + exz + x3y = 1.
(a) Near which points can we describe S as the graph
of a C1 function z = f(x, y)? What is f{x, y) in
this case?
(b) Describe the set of "bad" points of S, that is, the
points (xo, yo, zo) € S where we cannot describe
S as the graph of a function z = f(x, y).
(c) Use a computer to help give a complete picture of
S.
42. Let F(x, y) = c define a curve C in R2. Suppose
(xo, yo) is a point ofC such that VF(xo, yo) / 0. Show
that the curve can be represented near (xo, yo) as either
the graph of a function y = f(x) or the graph of a
function x = g(y).
43. Let F(x, y) = x2 — y3, and consider the curve C de-
fined by the equation F(x, y) = 0.
(a) Show that (0, 0) lies on C and that Fy(0, 0) = 0.
(b) Can we describe C as the graph of a function
y = /(x)? Graph C.
(c) Comment on the results of parts (a) and (b) in light
of the implicit function theorem (Theorem 6.5).
44. (a) Consider the family of level sets of the function
F(x, y) = xy + 1. Use the implicit function theo-
rem to identify which level sets of this family are
actually unions of smooth curves in R2 (i.e., locally
graphs of C1 functions of a single variable).
(b) Now consider the family of level sets of
F(x, y, z) = xyz + 1. Which level sets of this
family are unions of smooth surfaces in R3?
45. Suppose that F(u, v) is of class C1 and is such that
F(-2, 1) = 0 and F„(-2, 1) = 7, F„(-2, 1) = 5. Let
G(x, y, z) = F(x3 - 2y2 + z5, xy - x2z + 3).
(a) Check that G(-l, 1, 1) = 0.
(b) Show that we can solve the equation G(x, y, z) =
0 for z in terms of x and y (i.e., as z = g(x, y), for
(x,y)near(-l, 1) so that g(-l, 1)= 1).
46. Can you solve
X2j2 — x\ cos yi = 5
X2 sin yi + x\yi = 2
for yi, y2 as functions of x\, xi near the point
(x\, X2, yu y2) = (2, 3, n, 1)? What about near the
point (xi , X2, yi , ya) = (0, 2, n/2, 5/2)?
Chapter 2 | Differentiation in Several Variables
x\y\
47. Consider the system
2x2y3 = 1
xiy{ + x2y2 - 4y2yi = -9 .
x2y\ + 3xiyj = 12
(a) Show that, near the point (x\, x%, y\, y%, y{) =
(1, 0, —1, 1, 2), it is possible to solve for yi, yj,
ys in terms of x\, x%.
(b) From the result of part (a), we may consider y\,y%,
j3 to be functions of.ti and.*^. Use implicit differ-
dy\
entiation and the chain rule to evaluate (1, 0),
dx\
-i(l,0),and-^(l,0).
9xi dx\
48. Consider the equations that relate cylindrical and
Cartesian coordinates in R3:
x = r cos 8
y = r sin 8 .
z = z
(a) Near which points of R3 can we solve for r, 8, and
z in terms of the Cartesian coordinates?
(b) Explain the geometry behind your answer in
part (a).
49. Recall that the equations relating spherical and Carte-
sian coordinates in R3 are
' x = p sin <p cos 9
y = p sinip sin#.
Z = p cos <p
(a) Near which points of R3 can we solve for p, <p, and
8 in terms of x, y, and z?
(b) Describe the geometry behind your answer in
part (a).
Figure 2.75 The tangent line to
y = f(x) at (xq, f(xo)) crosses the
x-axis at x = x\.
2.7 Newton's Method (optional)
When you studied single-variable calculus, you may have learned a method, known
as Newton's method (or the Newton-Raphson method), for approximating the
solution to an equation of the form f(x) = 0, where /: X C R —> R is a differ-
entiable function. Here's a reminder of how the method works.
We wish to find a number r such that f(r ) = 0. To approximate r, we make
an initial guess xq for r and, in general, we expect to find that f(xo) ^ 0. So next
we look at the tangent line to the graph of / at (x0, f(x0)). (See Figure 2.75.)
Since the tangent line approximates the graph of / near (xo, f(xo)), we can
find where the tangent line crosses the x-axis. The crossing point (jci, 0) will
generally be closer to (r, 0) than (xq, 0) is, so we take x\ as a revised and improved
approximation to the root r of f{x) = 0.
To find x\ , we begin with the equation of the tangent line
y = f(*o) + /'C*o)(* - *o),
then set y = 0 to find where this line crosses the x -axis. Thus, we solve the equation
/(*o) + f'(x0)(xi - x0) = 0
for x\ to find that
f(xo)
X] — Xr> .
f'(x0)
Once we have x\, we can start the process again using x\ in place of x0 and
produce what we hope will be an even better approximation x^ via the formula
fix,)
Xi = X\ .
f'(xi)
Indeed, we may iterate this process and define Xk recursively by
f{Xk-\)
xt = Xk-i k = 1,2,...
and thereby produce a sequence of numbers xq, x\, . . . , Xk,
(1)
2.7 | Newton's Method (optional) 177
It is not always the case that the sequence {xk\ converges. However, when
it does, it must converge to a root of the equation f(x) = 0. To see this, let
L = lim^oo Xk. Then we also have lim^oo jc&_i = L. Taking limits in formula
(1), we find
L=L-m,
/'(!)'
which immediately implies that f(L) = 0. Hence, L is a root of the equation.
Now that we have some understanding of derivatives in the multivariable
case, we turn to the generalization of Newton's method for solving systems of n
equations in n unknowns. We may write such a system as
fi(xi, ...,x») = 0
fi(x\, x„) = 0
. ■ (2)
f„(x\, ...,*„) = 0
We consider the map f: X c R" -> R" defined as f(x) = (/i(x), . . . , /n(x)) (i.e.,
f is the map whose component functions come from the equations in (2). The
domain X of f may be taken to be the set where all the component functions are
denned.) Then to solve system (2) means to find a vector r = (rj , . . . , r„) such
that f(r) = 0. To approximate such a vector r, we may, as in the single- variable
case, make an initial guess xo for what r might be. If f is differentiable, then we
know that y = f(x) is approximated by the equation
y = f(x<,) + Of(x0)(x - xo).
(Here we think of f(xo) and the vectors x and xo as n x 1 matrices.) Then we set
y equal to 0 to find where this approximating function is zero. Thus, we solve the
matrix equation
f(x„) + £>f(x0)(x1 - x0) = 0 (3)
for xi to give a revised approximation to the root r. Evidently (3) is equivalent to
Df(xo)(Xl - x0) = -f(x0). (4)
To continue our argument, suppose that Df(xo) is an invertible n x n matrix,
meaning that there is a second n x n matrix [Df(xo)]_1 with the property that
[Df^r'Dftxo) = JDf(x0)[Df(x0)]"1 = /„, the n x n identity matrix. (See Ex-
ercises 20 and 30-38 in §1.6.) Then we may multiply equation (4) on the left by
[Df(x0)]_1 to obtain
7„(X! -x0)=-[Df(x0)r1f(x0).
Since I„A = A for any n x k matrix A, this last equation implies that
Xl =x0-[£>f(x0)r1f(x0). (5)
As we did in the one -variable case of Newton's method we may iterate formula
(5) to define recursively a sequence {x^} of vectors by
xt = xn-[Df(Xn)r'%-i) (6)
Chapter 2 | Differentiation in Several Variables
-3 +
Figure 2.76 Finding
the intersection
points of the circle
4 and
the hyperbola
Ax2 - y2 = A in
Example 1.
Note the similarity between formulas (1) and (6). Moreover, just as in the case
of formula (1), although the sequence {xo, Xi, . . . , xk, . . .} may not converge, if
it does, it must converge to a root of f(x) = 0. (See Exercise 4.)
EXAMPLE 1 Consider the problem of finding the intersection points of the cir-
cle x2 + y2 = 4 and the hyperbola 4x2 — y2 = 4. (See Figure 2.76.) Analytically,
we seek simultaneous solutions to the two equations
x2 + y2 = 4 and Ax2 - y2 = 4,
or, equivalently, solutions to the system
x2 + y2 - 4 = 0
Ax2 - y2
0
(7)
To use Newton's method, we define a function f : R -> R by f(x , y) = (x +
y
A Ax1
A) and try to approximate solutions to the vector equation
f(x, y) = (0, 0). We may begin with any initial guess, say,
x0
Xo
T
l
and then produce successive approximations Xi, x2,
mula (6). In particular, we have
to a solution using for-
NotethatdetDf(x, y)
2x
8.v
2y
-2y
Df(x,y) =
-20xy . You may verify (see Exercise 36 in § 1 .6) that
[Df(x,y)Tl =
1
-20xy
-2y -2y
-8x 2x
1
2
57
1
1
Thus,
~xk~
~Xk-\
yk-\_
~xk-\
yk-i
[Df(xk_ uyk-i )] f(*k- 1 . yk-i)
l l
2
L 5yk_
1
10%-i J
4-i + yLi-4
^U-yU-A
Jxk-\
L I0yk-i J
Xk-l
yk-\ -
JXk-\
lOxjt i
10W-i J
2.7 | Newton's Method (optional) 179
Beginning with xo = yo = 1 , we have
5 ■ l2 — 8 5 • l2 — 12
xx = \ = 1.3 yi = 1 = 1.7
10-1 10-1
5(1.3)2-8 5(1.7)2 - 12
x2 = 1.3 = 1.265385 y2 = 1.7
10(1.3) y 10(1.7)
= 1.555882, etc.
It is also easy to hand off the details of the computation to a calculator or a
computer. One finds the following results:
k
xk
y*
0
1
l
1
1.3
1.7
2
1.26538462
1.55588235
3
1.26491115
1.54920772
4
1.26491106
1.54919334
5
1.26491106
1.54919334
Thus, it appears that, to eight decimal places, an intersection point of the curves
is (1.26491106, 1.54919334).
In this particular example, it is not difficult to find the solutions to (7) exactly.
We add the two equations in (7) to obtain
5xz - 8 = 0
x2 =
Thus, x = ±V8/5. If we substitute these values for x into the first equation of
(7), we obtain
I + r ~ 4 = 0 y
Hence, y = ±y/l2/5. Therefore, the four intersection points are
11
5 ■
Since 78/5 « 1.264911064 and ^/T2/5 « 1.54919334, we see that Newton's
method provided us with an accurate approximate solution very quickly. ♦
EXAMPLE 2 We use Newton's method to find solutions to the system
x3 - 5x2 + 2x - y + 13 = 0
x3 +x2 - 14x - y - 19 = 0
(8)
As in the previous example, we define f : R2
y + 13, x3 + x2 - 14x - y - 19). Then
R2byf(x, y) = (x3 - 5x2 + 2x
Df(x,y) =
3x2 - lOx + 2
3x2 + 2x - 14
1 80 Chapter 2 | Differentiation in Several Variables
so that det Di(x, y) = \2x — 16 and
1
1
[Df(x,y)Y
Thus, formula (6) becomes
1
I2x - 16
-3x2 - 2x + 14
12x - 16
xk
~xk-\
\2x - 16
-3x2 -lOx + 2
\2x - 16
1
12xt.
-3x
16
2*fc_i + 14
-3x
12xt.
2
i - 16
lOjCjt-i +2
12x^-1 — 16
5x?_, + 2xk-\ -
12xt_
<r-l
16
+ X
k-1
I4xk-
y*-i + 13
y*-i - 19
6r2
I6xk-i — 32
3.v
16.T
\Ax\_x
\2xk-\ — 16
82xt_i - 8y*_i + 6jCjt_iyjt_i + 72
6^yt-l — 8
This is the formula we iterate to obtain approximate solutions to (8).
If we begin with x0 = (x0, y0) = (8, 10), then the successive approximations
\k quickly converge to (4, 5), as demonstrated in the table below.
k
Xk
yk
0
8
10
1
5.2
-98.2
2
4.1862069
-2.7412414
3
4.00607686
4.82161865
4
4.00000691
4.99981073
5
4.00000000
5.00000000
6
4.00000000
5.00000000
If we begin instead with x0 = (50, 60), then convergence is, as you might predict,
somewhat slower (although still quite rapid):
k
Xk
yk
0
50
60
1
25.739726
-57257.438
2
13.682211
-7080.8238
3
7.79569757
-846.58548
4
5.11470969
-86.660453
5
4.1643023
-1.6486813
6
4.00476785
4.86119425
7
4.00000425
4.99988349
8
4.00000000
5.00000000
9
4.00000000
5.00000000
2.7 I Exercises 181
On the other hand, if we begin with xo = (—2, 12), then the sequence of points
generated converges to a different solution, namely, (—4/3, —25/27):
k
Xk
yk
0
-2
12
1
-1.4
1.4
2
-1.3341463
-0.903122
3
-1.3333335
-0.9259225
4
-1.3333333
-0.9259259
5
-1.3333333
-0.9259259
In fact, when a system of equations has multiple solutions, it is not always
easy to predict to which solution a given starting vector xo will converge under
Newton's method (if, indeed, there is convergence at all). ♦
Finally, we make two remarks. First, if at any stage of the iteration process the
matrix Df(xk) fails to be invertible (i.e., [Df(x,t)]_1 does not exist), then formula
(6) cannot be used. One way to salvage the situation is to make a different choice
of initial vector xo in the hope that the sequence {x^} that it generates will not
involve any noninvertible matrices. Second, we note that if, at any stage, x,t is
exactly a root of f(x) = 0, then formula (6) will not change it. (See Exercise 7).
2.7 Exercises
1 . Use Newton's method with initial vector xo = (1 , — 1)
to approximate the real solution to the system
y V = 3
2yex + 10y4 = 0 '
2. In this problem, you will use Newton's method to
estimate the locations of the points of intersection
of the ellipses having equations 3x2 + y2 = 7 and
x2 +4y2 = 8.
(a) Graph the ellipses and use your graph to give a very
rough estimate (xo, yo) of the point of intersection
that lies in the first quadrant.
(b) Denote the exact point of intersection in the first
quadrant by (X, Y). Without solving, argue that
the other points of intersection must be (—X, Y),
(X, -Y), and (—X, —Y).
(c) Now use Newton's method with your estimate
(xq, yo) in part (a) to approximate the first quadrant
intersection point (X, Y).
(d) Solve for the intersection points exactly, and com-
pare your answer with your approximations.
3. This problem concerns the determination of the points
of intersection of the two curves with equations jc3 —
4y3 = 1 and x2 + 4y2 = 2.
(a) Graph the curves and use your graph to give rough
estimates for the points of intersection.
(b) Now use Newton's method with different initial
estimates to approximate the intersection points.
4. Consider the sequence of vectors xo, Xi, . . . , where,
for k > 1, the vector x* is defined by the Newton's
method recursion formula (6) given an initial "guess"
xo at a root of the equation f(x) = 0. (Here we as-
sume that f:XC R" — > R" is a differentiable func-
tion.) By imitating the argument in the single-variable
case, show that if the sequence {xk} converges to a vec-
tor L and Z)f(L) is an invertible matrix, then L must
satisfy f(L) = 0.
5. This problem concerns the Newton's method iteration
in Example 1 .
(a) Use initial vector xo = (— 1, 1) and calculate the
successive approximations Xi, X2, X3, etc. To what
solution of the system of equations (7) do the ap-
proximations converge?
(b) Repeat part (a) with xo = (1, — 1). Repeat again
with xo = (— 1, —1).
(c) Comment on the results of parts (a) and (b) and
whether you might have predicted them. Describe
the results in terms of Figure 2.76.
1 82 Chapter 2 | Differentiation in Several Variables
8. Suppose that f:XCR2->R2 is differentiable and
that we write f(x, y) = (f(x, y), g(x, y)). Show that
formula (6) implies that, for k > 1,
f(xk-i, yk-i)gy(xk-U Vk-l) ~ g(Xk-l, yk-\)fy(xk-\, yk-i)
fx(xk-i, yk-i)gy{xk-i, jh) - fy(xk-u yk-i)gx(xk-u yk-\)
g(xk_i,yk_\)fx{xk-\, yk-i) - /fa-i, yk-i)gx(xk-U yk-i)
fx(xk-u yk-i)gy(xk-u yk-i) - fy(xk-u yk-\)gx(xk-u yk-i)
6. Consider the Newton's method iteration in Example 2.
(a) Use initial vector xo = (1.4, 10) and calculate the
successive approximations xi, X2, X3, etc. To what
solution of the system of equations (8) do the ap-
proximations converge?
^ (b) Repeat part (a) with x0 = (1.3, 10).
(c) In Example 2 we saw that (4, 5) was a solution of
the given system of equations. Is (1.3, 10)closerto
(4, 5) or to the limiting point of the sequence you
calculated in part (b)?
(d) Comment on your observations in part (c). What
do these observations suggest about how easily you
can use the initial vector xo to predict the value of
lim^oo xk (assuming that the limit exists)?
7. Suppose that at some stage in the Newton's method it-
eration using formula (6), we obtain a vector xk that is
an exact solution to the system of equations (2). Show
that all the subsequent vectors x^+i, Xk+2, ■ • ■ are equal
to X£. Hence, if we happen to obtain an exact root via
Newton's method, we will retain it.
9. As we will see in Chapter 4, when looking for maxima
and minima of a differentiable function F:XCR"^
R, we need to find the points where DF{x\ , . . . , xn) =
[0 • • • 0], called critical points of F. Let F(x, y) =
4 sin (xy) + x3 + y 3 . Use Newton's method to approx-
imate the critical point that lies near (x , y) = (—1, —1).
1 0. Consider the problem of finding the intersection points
of the sphere x2 + y2 + 7} = 4, the circular cylinder
x2 + y2 = 1, and the elliptical cylinder 4y2 + z2 = 4.
(a) Use Newton's method to find one of the intersec-
tion points. By choosing a different initial vector
xo = (xq, yo, zo), approximate a second intersec-
tion point. (Note: You may wish to use a computer
algebra system to determine appropriate inverse
matrices.)
(b) Find all the intersection points exactly by means of
algebra and compare with your results in part (a).
True/False Exercises for Chapter 2
1 . The component functions of a vector- valued function 8.
are vectors.
2. The domain of f(x, y) = |^x + y + 1
{(x,y)GR2 | y /0,x # v}.
3 x
3. The range of f(x, y) = yx2 + yL + 1
{(u, v, w) G R3 | u > 1}.
x + y y
3 x
x + y y
4. The function f: R3 - {(0, 0, 0)} -> R3, f(x) = 2x/||x||
is one-one.
5. The graph of x = 9y2 + z2/4 is a paraboloid.
6. The graph of z + x2 = y2 is a hyperboloid.
The graph of any function of two variables is a level set of
a function of three variables.
The level set of any function of three variables is the graph
of a function of two variables.
10.
lim
2v2
(x,v)^(0,0) x2 + y2
1.
11. If/(x,y)
4 4
y — x
x2 + y2
2 when (x,y) = (0,0)
when(x,y)/(0,0\ then / is
continuous.
12.
7. The level set of a function f(x,y,z) is either empty 13.
or a surface.
If f(x,y) approaches a number L as (x, y) — > (a, b)
along all lines through (a,b), then lim(x y)^(aj,)f(x,y)
= L.
If limx^a f(x) exists and is finite, then f is continuous
at a.
Miscellaneous Exercises for Chapter 2
14. fx(a,b)= lim
/(x, 5) - /(a, fc)
15. If f(x, y, z) = siny, then V/(x, y, z) = cosy.
16. If f:R3 -* R4 is differentiable, then Df(x) is a 3 x 4
matrix.
17. Iff is differentiable at a, then f is continuous at a.
18. If f is continuous at a, then f is differentiable at a.
1 9. If all partial derivatives 3 f/dx\ , . . . , 3// 3x„ of a func-
tion f(xi , . . . , xn) exist at a = (ai , . . . , a„), then / is
differentiable at a.
20. Iff: R4 -+ R5 and g: R4 -+ R5 are both differentiable
at a G R4, then D(f- g)(a) = Df(a) - Dg(a).
21. There's a function / of class C2 such that
3/ i 3/ 7
— = y — 2x and — = y — 3xy .
ox dy
22. If the second-order partial derivatives of / exist at
(a, b), then fxy(a, b) = fyx(a, b).
23. If w = F(x, y, z) and z = g(x, y) where F and g are
differentiable, then
dw dF dF dg
dx dx dz dx
24. The tangent plane to z = x1 /(y + 1) at the point
(—2, 0, —8) has equation z = 12x + 8y + 16.
25. The plane tangent to xy/z2 = 1 at (2, 8, —4) has equa-
tion 4x + y + 2z = 8.
26. The plane tangent to the surface x2 + xyez + y3 =
1 at the point (2,-1,0) is parallel to the vector
3i+5j-3k.
3/
27. Dj/(x,y,Z):
28. D.kf(x, y, z)
dy
3/
dz'
29. If f(x, y) = sinx cosy and v is a unit vector in R2,
thenO^Dv/^-,-) < — .
30. If v is a unit vector in R3 and f(x, y, z) = sinx —
cosy + sinz, then
-V3 < Dyf(x,y,z) < <&.
Miscellaneous Exercises for Chapter 2
1. Letf(x) = (i + k) x x.
(a) Write the component functions of f.
(b) Describe the domain and range of f.
order. Complete the following table by matching each
function in the table with its graph and plot of its level
curves.
Graph
Level curves
2. Let f(x) = proj3i_2j+kx, where x = xi + yj + zk.
Function
(uppercase
(lowercase
(a) Describe the domain and range of f.
fix, y)
letter)
letter)
(b) Write the component functions of f.
f(*,y) =
1
x2 + y2 + 1
f(x, y) =
sin y/x2 + y2
3. Let f(x, y) = ^/xy.
fix, y) =
(3y2 - 2x2)e--r2-2>'2
(a) Find the domain and range of /.
fix, y) =
3 1 2
y — 3x y
(b) Is the domain of / open or closed? Why?
fix, y) =
x2y2e-x*-2?
fix, y) =
-x2-v2
ye y
4. Letg(x, y) =
(a) Determine the domain and range of g.
(b) Is the domain of g open or closed? Why?
5. Figure 2.77 shows the graphs of six functions /(x, y)
and plots of the collections of their level curves in some
6. Consider the function /(x, y) = 2 + ln(x2 + y2).
(a) Sketch some level curves of /. Give at least those
at heights, 0, 1, and 2. (It will probably help if you
give a few more.)
(b) Using part (a) or otherwise, give a rough sketch of
the graph of z = f(x, y).
1 84 Chapter 2 | Differentiation in Several Variables
J
1
1
-2-10 1 2
Figure 2.77 Figures for Exercise 5.
7. Use polar coordinates to evaluate
,2
lim
yx -y
(x,y)^{0,0) x2 + y2
8. This problem concerns the function
i 2xy
f(x,y)--
if(x,y)
C2 + V2
0
if(x,y) ^(0,0)
(0,0)
(a) Use polar coordinates to describe this function.
(b) Using the polar coordinate description obtained in
part (a), give some level curves for this function.
(c) Prepare a rough sketch of the graph of /.
(d) Determine lim(A: ,,)^(o,o) f(x , y), if it exists.
(e) Is / continuous? Why or why not?
9. Let
xy(xy + x2)
F(x, y)
if(x,y)#(0,0)
x4 + y4
0 if (x, y) = (0, 0)
10.
Show that the function g(x ) = F (x , 0) is continuous at
x = 0. Show that the function h(y) = F(0, y) is con-
tinuous at y = 0. However, show that F fails to be
continuous at (0, 0). (Thus, continuity in each variable
separately does not necessarily imply continuity of the
function.)
Suppose /: U c R" R is not defined at a point
a G R" but is defined for all x near a. In other words,
the domain U of / includes, for some r > 0, the set
Br = {x G R" | 0 < || x - a|| < r\. (The set Br is just
an open ball of radius r centered at a with the point
Miscellaneous Exercises for Chapter 2
a deleted.) Then we say limx^a /W = +00 if f(x)
grows without bound as x -> a. More precisely, this
means that given any N > 0 (no matter how large),
there is some S > 0 such that if 0 < ||x — a|| < S (i.e.,
ifx € fi,-), then /(a) > N.
(a) Using intuitive arguments or the preceding tech-
nical definition, explain why lim^o l/x2 = 00.
(b) Explain why
2
lim r = 00.
(x,y)^(l,3) (x - l)2 + (y - 3)2
(c) Formulate a definition of what it means to say that
lim /(x) = —00.
x— *a
(d) Explain why
1 -x
lim — ; = —00.
(x,j.)^(0,0) xy4 - y4 + x3 - x2
Exercises 11—1 7 involve the notion of windchill temperature —
see Example 7 in §2. 1, and refer to the table of windchill values
on page 85.
11. (a) Find the windchill temperature when the air tem-
perature is 25 °F and the windspeed is 10 mph.
(b) If the windspeed is 20 mph, what air temperature
causes a windchill temperature of — 1 5 °F?
12. (a) If the air temperature is 1 0 °F, estimate (to the near-
est unit) what windspeed would give a windchill
temperature of —5 °F.
(b) Do you think your estimate in part (a) is high or
low? Why?
13. At a windspeed of 30 mph and air temperature of 35 °F,
estimate the rate of change of the windchill tempera-
ture with respect to air temperature if the windspeed is
held constant.
14. At a windspeed of 1 5 mph and air temperature of 25 °F,
estimate the rate of change of the windchill tempera-
ture with respect to windspeed.
15. Windchill tables are constructed from empirically de-
rived formulas for heat loss from an exposed sur-
face. Early experimental work of P. A. Siple and C. F.
Passel,4 resulted in the following formula:
W = 91 .4 + (t - 91 .4)(0.474 + 0.304V? - 0.0203s).
Here W denotes windchill temperature (in degrees
Fahrenheit), t the air temperature (for t < 9 1 .4 °F), and
s the windspeed in miles per hour (for s > 4 mph).5
(a) Compare your answers in Exercises 1 1 and 12 with
those computed directly from the Siple formula
just mentioned.
(b) Discuss any differences you observe between your
answers to Exercises 1 1 and 12 and your answers
to part (a).
(c) Why is it necessary to take t < 91.4°F and
s > 4 mph in the Siple formula? (Don't look for
a purely mathematical reason; think about the
model.)
1 6. Recent research led the United States National Weather
Service to employ a new formula for calculating wind-
chill values beginning November 1, 2001. In partic-
ular, the table on page 85 was constructed from the
formula
W = 35.74 + 0.621f - 35.75j016 + 0.4275^016.
Here, as in the Siple formula of Exercise 15, W de-
notes windchill temperature (in degrees Fahrenheit),
t the air temperature (for t < 50 °F), and s the wind-
speed in miles per hour (for s > 3 mph).6 Compare
your answers in Exercises 13 and 14 with those com-
puted directly from the National Weather Service
formula above.
17. In this problem you will compare graphically the two
windchill formulas given in Exercises 15 and 16.
(a) If W\ (s, t) denotes the windchill function given by
the Siple formula in Exercise 15 and 0 the
windchill function given by the National Weather
Service formula in Exercise 16, graph the curves
y = Wi(s, 40) and y = Wi(s, 40) on the same set
of axes. (Let s vary between 3 and 120 mph.) In
addition, graph other pairs of curves y = W\(s, to),
y = W2(s, ?o) for other values of to. Discuss what
your results tell you about the two windchill
formulas.
(b) Now graph pairs of curves y = Wi(sq, t), y =
W2CS0, t) for various constant values so for wind-
speed. Discuss your results.
(c) Finally, graph the surfaces z = W\{s, t) and
z = W2(s, t) and comment.
"Measurements of dry atmospheric cooling in sub freezing temperatures," Proc. Amer. Phil. Soc, 89
(1945), 177-199.
5 From Bob Rilling, Atmospheric Technology Division, National Center for Atmospheric Research
(NCAR), "Calculating Windchill Values," February 12, 1996. Found online at http://www.atd.ucar.edu/
homes/rilling/wc_formula.html (July 31, 2010).
6 NOAA, National Weather Service, Office of Climate, Water, and Weather Services, "NWS Wind Chill
Temperature Index." February 26, 2004. <http://www.nws.noaa.gov/om/ windchill> (July 31, 2010).
1 86 Chapter 2 | Differentiation in Several Variables
1 8. Consider the sphere of radius 3 centered at the origin.
The plane tangent to the sphere at (1, 2, 2) intersects
the x-axis at a point P. Find the coordinates of P.
1 9. Show that the plane tangent to a sphere at a point P on
the sphere is always perpendicular to the vector OP
from the center O of the sphere to P. (Hint: Locate the
sphere so its center is at the origin in R3.)
20. The surface z = 3x2 + |x3 — ^x4 — 4y2 is inter-
sected by the plane 2x — y = I. The resulting intersec-
tion is a curve on the surface. Find a set of parametric
equations for the line tangent to this curve at the point
21. Consider the cone z2 = x2 + y2.
(a) Find an equation of the plane tangent to the cone
at the point (3, -4, 5).
(b) Find an equation of the plane tangent to the cone
at the point (a, b, c).
(c) Show that every tangent plane to the cone must
pass through the origin.
22. Show that the two surfaces
Si: z = xy and S2: z = \x2 — y
intersect perpendicularly at the point (2, 1,2).
23. Consider the surface z = x2 + Ay2.
(a) Find an equation for the plane that is tangent to the
surface at the point (1, —1,5).
(b) Now suppose that the surface is intersected with the
plane x = 1 . The resulting intersection is a curve
on the surface (and is a curve in the plane x = 1
as well). Give a set of parametric equations for the
line in R3 that is tangent to this curve at the point
(1 , — 1 , 5). A rough sketch may help your thinking.
24. A turtleneck sweater has been washed and is now tum-
bling in the dryer, along with the rest of the laundry. At
a particular moment to, the neck of the sweater mea-
sures 1 8 inches in circumference and 3 inches in length.
However, the sweater is 100% cotton, so that at to the
heat of the dryer is causing the neck circumference to
shrink at a rate of 0.2 in/min, while the twisting and
tumbling action is causing the length of the neck to
stretch at the rate of 0.1 in/min. How is the volume V
of the space inside the neck changing at t = ?o? Is V
increasing or decreasing at that moment?
25. A factory generates air pollution each day according
to the formula
P(S, T) = 33052/3T4/5,
where S denotes the number of machine stations in
operation and T denotes the average daily tempera-
ture. At the moment, 75 stations are in regular use and
the average daily temperature is 15 °C. If the average
temperature is rising at the rate of 0.2°C/day and the
number of stations being used is falling at a rate of
2 per month, at what rate is the amount of pollution
changing? (Note: Assume that there are 24 workdays
per month.)
26. Economists attempt to quantify how useful or satisfy-
ing people find goods or services by means of utility
functions. Suppose that the utility a particular individ-
ual derives from consuming x ounces of soda per week
and watching y minutes of television per week is
u(x, y) = 1
-0.001.v2-0.00005v2
Further suppose that she currently drinks 80 oz of soda
per week and watches 240 min of TV each week. If she
were to increase her soda consumption by 5 oz/week
and cut back on her TV viewing by 15 min/week, is
the utility she derives from these changes increasing
or decreasing? At what rate?
27. Suppose that w = x2 + y2 + z2 andx = p cos# s'mcp,
y = p sin 6 sin cp, z = p cos cp . (Note that the equations
for x, y, and z in terms of p, cp, and 6 are just the con-
version relations from spherical to rectangular coordi-
nates.)
(a) Use the chain rule to compute dw/dp, dw/d(p,
and dw/dd. Simplify your answers as much as
possible.
(b) Substitute p, (p, and 9 for x, y, and z in the original
expression for w. Can you explain your answer in
part (a)?
28. If w = f (—t-^)' show that
,3w ,3iu
x2 y2 — = 0.
3x dy
(You should assume that / is a differentiable function
of one variable.)
29. Let z = g(x, y) be a function of class C2, and let
x = er cos#, v = er sin#.
(a) Use the chain rule to find dz/dr and 3 z/ 9 6 in terms
of dz/dx and dz/dy. Use your results to solve for
3z/3x and dz/dy in terms of dz/dr and dz/d0.
(b) Use part (a) and the product rule to show that
32z 32z _ _2r {dh dh
3x2 + ^2 - e [dr2 + d92
30. (a) Use the function /(x, y) = xy (= eylnx) and the
d
multivariable chain rule to calculate — (w").
du
(b) Use the multivariable chain rule to calculate
^((sinOcos').
dt
Miscellaneous Exercises for Chapter 2
31. Use the function f(x, y, z) = xy andthemultivariable
d , «,
chain rule to calculate — (u ).
du
32. Suppose that /: R" — » R is a function of class C2. The
Laplacian of /, denoted V2/, is defined to be
V2/
92/ a2/
— — -i — — + ■
dx} dx2
+
a2/
dx2'
When n = 2 or 3, this construction is important when
studying certain differential equations that model phys-
ical phenomena, such as the heat or wave equations.
(See Exercises 28 and 29 of §2.4.) Now suppose that
/ depends only on the distance x = (x\, . . . , x„) is
from the origin in R" ; that is, suppose that /(x) = g(r)
for some function g, where r = ||x|| . Show that for all
x / 0, the Laplacian is given by
V2/
g'(r) + g"(r).
33. (a) Consider a function f(x, y) of class C4. Show
that if we apply the Laplacian operator V2 =
d2/dx2 + d2/dy2 twice to /, we obtain
V2(V2/)
-L + 2 -
dx4 dx2dy2
+
9V
3y4 '
(b) Now suppose that / is a function of « variables of
class C4. Show that
V2(V2/)=£
94/
, dxfdx1-
',7 = 1 ' J
Functions that satisfy the partial differential equa-
tion V2(V2/) = 0 are called Inharmonic func-
tions and arise in the theoretical study of elasticity.
34. Livinia, the housefly, finds herself caught in the oven
at the point (0, 0, 1). The temperature at points in the
oven is given by the function
T(x,y,z)= \0(xe->2 +ze~xl),
where the units are in degrees Celsius.
(a) IfLivinia begins to move toward the point (2, 3, 1),
at what rate (in deg/cm) does she find the temper-
ature changing?
(b) In what direction should she move in order to cool
off as rapidly as possible?
(c) Suppose that Livinia can fly at a speed of 3 cm/sec.
If she moves in the direction of part (b), at what
(instantaneous) rate (in deg/sec) will she find the
temperature to be changing?
35. Consider the surface given in cylindrical coordinates
by the equation z = r cos 36.
(a) Describe this surface in Cartesian coordinates, that
is, asz = f{x,y).
(b) Is / continuous at the origin? (Hint: Think cylin-
drical.)
(c) Find expressions for df /dx and df/dy at points
other than (0, 0). Give values for df /dx and df/ dy
at (0, 0) by looking at the partial functions of /
through (x, 0) and (0, y) and taking one-variable
limits.
(d) Show that the directional derivative Duf(0, 0) ex-
ists for every direction (unit vector) u. (Hint: Think
in cylindrical coordinates again and note that you
can specify a direction through the origin in the
xy-plane by choosing a particular constant value
for 9.)
(e) Show directly (by examining the expression for
df/dy when(jc, y) / (0, 0) and also using part (c))
that df/dy is not continuous at (0, 0).
(f) Sketch the graph of the surface, perhaps using a
computer to do so.
36. The partial differential equation
d2u d2u d2u
Jx2 + Jv2 + Jz2 ~C~dt2
d2u
is known as the wave equation. It models the motion
of a wave u ( x , y , z , t ) in R3 and was originally derived
by Johann Bernoulli in 1727. In this equation, c is a
positive constant, the variables x, y, and z represent
spatial coordinates, and the variable t represents time.
(a) Let u = cos(x — t) + sin(x + t) — 2ez+l — (y —
f)3 • Show that u satisfies the wave equation with
c= 1.
(b) More generally, show that if f\, fa, g\, g%, hi, and
hi are any twice differentiable functions of a single
variable, then
u(x, y, z, t) = fi(x - t) + f2(x + t)
+ gi(y - 1) + g2(y + 1)
+ hl(z-t) + h2(z + t)
satisfies the wave equation with c = 1 .
Let X be an open set in R". A function F: X — > Ris said to be
homogeneous of degree d if, for all x = (x\, X2, ■ ■ ■ , xn) e X
and all t € R such that tx € X, we have
F(tx\ , tx2, . ■ ■ , tx„) = td F(x\ , x2, . . . , xn).
Exercises 37—44 concern homogeneous functions.
In Exercises 37-41, which of the given functions are homoge-
neous ? For those that are, indicate the degree dof homogeneity.
37. F(x,y) = x3 +xy2 - 6y3
38. F(x, y, z) = x3y - x2z2 + z&
39. F(x, y, z) = zy2 - x3 + x2z
1 88 Chapter 2 | Differentiation in Several Variables
40. F(x, y) = e
41. F(x,y,z)
JC3 + X2 v
yz'-
x\
z + Ixz2
42. If F(x, y, z) is a polynomial, characterize what it
means to say that F is homogeneous of degree d (i.e.,
explain what must be true about the polynomial if it is
to be homogeneous of degree d).
43. Suppose F{x\, X2, . . . , xn) is differentiable and homo-
geneous of degree d. Prove Euler's formula:
dF dF dF
xi- \-x2- 1 \-x„ — =dF.
OXl 0X2 OXn
(Hint: Take the equation F(tx\,tX2, ■ ■ ■ ,txn) =
td F(x\ , X2, . .. ,xn) that defines homogeneity and dif-
ferentiate with respect to / .)
44. Generalize Euler's formula as follows: If F is of class
C2 and homogeneous of degree d, then
d2F
f-r1. dxjdx:
Can you conjecture what an analogous formula
involving the fcth-order partial derivatives should look
like?
3.1 Parametrized Curves and
Kepler's Laws
3.2 Arclength and Differential
Geometry
3.3 Vector Fields: An
Introduction
3.4 Gradient, Divergence, Curl,
and the Del Operator
True/False Exercises for
Chapter 3
Miscellaneous Exercises
for Chapter 3
Z
y
X
Figure 3.1 The path x of
Example 1.
Vector-Valued
Functions
Introduction
The primary focus of Chapter 2 was on scalar-valued functions, although general
mappings from R" to Rm were considered occasionally. This chapter concerns
vector- valued functions of two special types:
1. Continuous mappings of one variable (i.e., functions x: / C R -+ R", where
/ is an interval, called paths in R").
2. Mappings from (subsets of) R" to itself (called vector fields).
An understanding of both concepts is required later, when we discuss line and
surface integrals.
3.1 Parametrized Curves and Kepler's Laws
Paths in Rn
We begin with a simple definition. Let / denote any interval in R. (So / can be
of the form [a, b], (a, b), [a, b), {a, b], [a, oo), (a, oo), (— oo, b], (—00, b), or
(—00, 00) = R.)
DEFINITION 1.1 A path in R" is a continuous function x: / -» R" . If / =
[a, b] for some numbers a <b, then the points x(a) and x(b) are called the
endpoints of the path x. (Similar definitions apply if / = [a, b), [a, 00), etc.)
EXAMPLE 1 Let a and b be vectors in R with a / 0. Then the function
x: (—00, 00) -> R3 given by
x(f) = b + ta
defines the path along the straight line parallel to a and passing through the end-
point of the position vector of b as in Figure 3.1. (See formula ( 1 ) of § 1 .2.) ♦
EXAMPLE 2 The path y: [0, 2tt) -> R2 given by
y(f) = (3 cosf, 3 sinf)
can be thought of as the path of a particle that travels once, counterclockwise,
around a circle of radius 3 (Figure 3.2). ♦
190 Chapter 3 I Vector-Valued Functions
Figure 3.4 The path x and its
velocity vector v.
EXAMPLE 3 The map z: R ->■ R3 defined by
z(f) = (a cos t, a sin?, bt), a, b constants {a > 0)
is called a circular helix, so named because its projection in the xy-plane is a
circle of radius a. The helix itself lies in the right circular cylinder x2 + y2 = a2
(Figure 3.3). The value of b determines how tightly the helix twists. ♦
We distinguish between a path x and its range or image set x(7), the latter
being a curve in R". By definition, a path is a function, a dynamic object (at least
when we imagine the independent variable / to represent time), whereas a curve
is a static figure in space. With such a point of view, it is natural for us to consider
the derivative Dx(t ), which we also write as x'(/ ) or v(/), to be the velocity vector
of the path. We can readily justify such terminology. Since
x(t) = (x1(t),x2(t),...,xn(t))
is a function of just one variable,
v(/) = x'(t) = lim
x(r + At) - x(t)
At '
Thus, v(/) is the instantaneous rate of change of position x(t) with respect to t
(time), so it can appropriately be called velocity. Figure 3.4 provides an indication
as to why we draw v(f) as a vector tangent to the path at x(f). Continuing in this
vein, we introduce the following terminology:
DEFINITION 1 .2 Let x: / ->■ R" be a differentiable path. Then the velocity
\(t) = x'(t) exists, and we define the speed of x to be the magnitude of
velocity; that is,
Speed = ||v(OI|.
If v is itself differentiable, then we call V(t) = x"(t) the acceleration of x
and denote it by m(t).
EXAMPLE 4 The helix x(t ) = (a cos t , a sin t , bt) has
v(/) = — a sinf i + a cost j + fok and a(f) = — a cos/ i — a sin/ j.
3.1 ; Parametrized Curves and Kepler's Laws
Thus, the acceleration vector is parallel to the xy-plane (i.e., is horizontal). The
speed of this helical path is
||v(OII = -vA-a sin t)2 + (a cos tf + b2 = y/a2 + b2,
which is constant. ♦
The velocity vector v is important for another reason, namely, for finding equations
of tangent lines to paths. The tangent line to a differentiate path x, at the point
x0 = x(?o), is the line through xo that is parallel to any (nonzero) tangent vector
to x at xo. Since v(t), when nonzero, is always tangent to x(t), we may use equa-
tion ( 1 ) of § 1 .2 to obtain the following vector parametric equation for the tangent
line:
l(s) = Xo + SV(). (1)
Here vo = v(to) and s may be any real number.
In equation (1), we have 1(0) = Xo. To relate the new parameter s to the
original parameter t for the path, we set s = t — to and establish the following
result:
PROPOSITION 1 .3 Let x be a differentiate path and assume that v0 =
v(?o) ^ 0. Then a vector parametric equation for the line tangent to x at x0 = x(t0)
is either
l(s) = x0 + s\0 (2)
or
1(0 = x0 + (t - f0)v0. (3)
(See Figure 3.5.)
EXAMPLE 5 If x(?) = (it + 2, t2 - 7, t - t2), we find parametric equations
for the line tangent to x at (5, — 6, 0) = x(l).
For this path, v(0 = x'(t) = 3i + 2/j + (1 - 2t)k, so that
Thus, by formula (3),
1(0
v0 = v(l) = 3i + 2j-k.
(5i-6j) + (f- l)(3i + 2j-k).
Taking components, we read off the parametric equations for the coordinates
of the tangent line as
x = 3t +2
y = 2t - 8 .
z=\-t
The physical significance of the tangent line is this: Suppose a particle of
mass m travels along a path x. If, suddenly, at t = to, all forces cease to act on the
particle (so that, by Newton's second law of motion F = ma, we have a(/) = 0
for t > to), then the particle will follow the tangent line path of equation (3).
192 Chapter 3 I Vector-Valued Functions
EXAMPLE 6 If Roger Ramjet is fired from a camion, then we can use vectors
to describe his trajectory. (See Figure 3.6.)
y
Roger's path
x
Figure 3.6 Roger Ramjet's path.
We'll assume that Roger is given an initial velocity vector vo by virtue of the
firing of the cannon and that thereafter the only force acting on Roger is due to
gravity (so, in particular, we neglect any air resistance). Let us choose coordinates
so that Roger is initially at the origin, and throughout our calculations we '11 neglect
the height of the cannon. Let x(r) = (x(f), y(t)) denote Roger's path. Then the
information we have is
a(t) = x"(t)=-g\
(i.e., the acceleration due to gravity is constant and points downward); hence,
v(0) = x'(0) = v0
and
x(0) = 0.
Since a(/) = V(t), we simply integrate the expression for acceleration compo-
nentwise to find the velocity:
v(0
= j a(t)dt = j -g]dr = -gtj +
Here c is an arbitrary constant vector (the "constant of integration"). Since v(0) =
vo, we must have c = vo, so that
v(0 = -gtj + v0.
Integrating again to find the path,
x(t) = j v(t)dt = j (-gt\ + \0)dt = -^gt2] + t\0 + d,
where d is another arbitrary constant vector. From the remaining fact that x(0) = 0,
we conclude that
1 7
x(0 = -^gn + ty0 (4)
describes Roger's path.
To understand equation (4) better, we write vo in terms of its components:
vo = vo cos 9 i + vo sin 9 j.
Here vo = ||vo|| is the initial speed. (We're really doing nothing more than
expressing the rectangular components of vo in terms of polar coordinates.
3.1 : Parametrized Curves and Kepler's Laws
See Figure 3.7.) Thus,
x(r) = — \gt2\ + f(i>0cos#i + vosin# j)
= (v0 cos 9)t\ + I (u0 sin 9)t — -gt2
1
2
From this, we may read off the parametric equations:
x = (vo cos6)t
1 , -
y = (v0 sinfl)/ - -gtz
from which it is not difficult to check that Roger's path traces a parabola. ♦
Here are two practical questions concerning the set-up of Example 6: First, for
a given initial velocity, how far does Roger travel horizontally? Second for a given
initial speed how should the cannon be aimed so that Roger travels (horizontally)
as far as possible? To find the range of the cannon shot and thereby answer the
first question, we need to know when y = 0 (i.e., when Roger hits the ground).
Thus, we solve
(u0 sin 6>)r - \gt2 = t(v0sm9 - \gt) = 0
for t. Hence, y = 0 when t = 0 (which is when Roger blasts off) and when
t = (2vq sin9)/g. At this later time,
(2vo sin#\ sin 29
— = ■ (5)
8 J S
Formula (5) is Roger's horizontal range for a given initial velocity. To maximize
the range for a given initial speed vo, we must choose 9 so that (v2 sin29)/g is
as large as possible. Clearly, this happens when sin 29 = 1 (i.e., when 9 = jt/4).
Kepler's Laws of Planetary Motion (optional)
Since classical antiquity, individuals have sought to understand the motions of
the planets and stars. The majority of the ancient astronomers, using a combina-
tion of crude observation and faith, believed all heavenly bodies revolved around
the earth. Fortunately, the heliocentric (or "sun-centered") theory of Nicholas
Copernicus (1473-1543) did eventually gain favor as observational techniques
improved. However, it was still believed that the planets traveled in circular or-
bits around the sun. This circular orbit theory did not correctly predict planetary
positions, so astronomers postulated the existence of epicycles, smaller circular
orbits traveling along the major circular arc, an example of which is shown in
Figure 3.8. Although positional calculations with epicycles yielded results closer
to the observed data, they still were not correct. Attempts at further improvements
were made using second- and third-order epicycles, but any gains in predictive
power were made at a cost of considerable calculational complexity. A new idea
was needed. Such inspiration came from Johannes Kepler (1571-1630), son of a
saloonkeeper and assistant to the Danish astronomer Tycho Brahe. The classical
astronomers were "stuck on circles" for they believed the circle to be a perfect
form and that God would use only such perfect figures for planetary motion.
Kepler, however, considered the other conic sections to be as elegant as the cir-
cle and so hypothesized the simple theory that planetary orbits are elliptical.
Empirical evidence bore out this theory.
194 Chapter 3 | Vector-Valued Functions
Figure 3.9 Kepler's second law
of planetary motion: If
t2 — t\ = t4 — f3, then A\ = A2,
where A; and A2 are the areas of
the shaded regions.
Kepler's three laws of planetary motion are
1. The orbit of a planet is elliptical, with the sun at a focus of the ellipse.
2. During equal periods of time, a planet sweeps through equal areas with respect
to the sun. (See Figure 3.9.)
3. The square of the period of one elliptical orbit is proportional to the cube of
the length of the semimajor axis of the ellipse.
Kepler's laws changed the face of astronomy. We emphasize, however, that
they were discovered empirically, not analytically derived from general physical
laws. The first analytic derivation is frequently credited to Newton, who claimed
to have established Kepler's laws (at least the first and third laws) in Book I of
his Philosophiae Naturalis Principia Mathematica (1687). However, a number of
scientists and historians of science now consider Newton's proof of Kepler's first
law to be flawed and that Johann Bernoulli (1667-1748) offered the first rigorous
derivation in 1710.1 In the discussion that follows, Newton's law of universal
gravitation is used to prove all three of Kepler's laws.
In our work below, we assume that the only physical effects are those be-
tween the sun and a single planet — the so-called two-body problem. (The n-body
problem, where n > 3 is, by contrast, an important area of current mathematical
research.) To set the stage for our calculations, we take the sun to be fixed at the
origin O in R3 and the planet to be at the moving position P. We also need the
following two "vector product rules," whose proofs we leave to you:
PROPOSITION 1.4
1. If x and y are differentiable paths in R", then
d dx dy
— (x • y) = v • h x • — .
dty J dt dt
2. If x and y are differentiable paths in R3, then
d dx dy
— (x xy)= — xy + xx — .
dty J dt J dt
First, we establish the following preliminary result:
PROPOSITION 1 .5 The motion of the planet is planar, and the sun lies in the
planet's plane of motion.
PROOF Let r = OP. Then r is a vector whose representative arrow has its tail
fixed at O. (Note that r = r(t); that is, r is a function of time.) If v = r'(f), we
will show that r x v is a constant vector c. This result, in turn, implies that r must
always be perpendicular to c and, hence, that r always lies in a plane with c as
normal vector.
To show that r x v is constant, we show that its derivative is zero. By part 2
of Proposition 1.4,
d dr dy
— (r xv)= — xv + rx — = vxv+rxa,
dt dt dt
For an indication of the more recent controversy surrounding Newton's mathematical accomplishments,
see R. Weinstock, "Isaac Newton: Credit where credit won't do," The College Mathematics Journal, 25
(1994), no. 3, 179-192, andC. Wilson, "Newton's orbit problem: A historian's response," Ibid., 193-200,
and related papers.
3.1 i Parametrized Curves and Kepler's Laws
by the definitions of velocity and acceleration. We know that vxv = 0 (why?), so
d
— (r x v) = r x a. (6)
dt
Now we use Newton's laws. Newton's law of gravitation tells us that the planet
is attracted to the sun with a force
GMm
F = — u, (7)
where G is Newton's gravitational constant (= 6.6720 x 10~n Nm2/kg2), M
is the mass of the sun, m is the mass of the planet (in kilograms), r = ||r||, and
u = r/ 1| i* || (distances in meters). On the other hand Newton's second law of
motion states that, for the planet,
F = ma.
Thus,
GMm
ma = r — u,
rL
or
GM
a = -r. (8)
Therefore, a is just a scalar multiple of r and hence is always parallel to r. In
view of equations (6) and (8), we conclude that
d
— (r xv) = rxa = 0
dt
(i.e., that r x v is constant).
THEOREM 1 .6 (Kepler's first law) In a two-body system consisting of one
sun and one planet, the planet's orbit is an ellipse and the sun lies at one focus of
that ellipse.
PROOF We will eventually find a polar equation for the planet's orbit and see
that this equation defines an ellipse as described. We retain the notation from
the proof of Proposition 1.5 and take coordinates for R3 so that the sun is at the
origin, and the path of the planet lies in the xy-plane. Then the constant vector
c = r x v used in the proof of Proposition 1.5 may be written as ck, where c is
some nonzero real number. This set-up is shown in Figure 3.10.
z
c = r x v
Figure 3.10 Establishing Kepler's laws.
196 Chapter 3 I Vector-Valued Functions
Step 1. We find another expression for c. By definition of u in formula (7),
r = ru, so that, by the product rule,
d du dr
v = — (ru) = r— + — u.
dt dt dt
Hence,
/ du dr \ 2 / du\ dr
c = r x v = (ru) x r 1 u =r ux — + r — (u x u).
\ dt dt J \ dt J dt
Since uxu must be zero, we conclude that
c = r'(uxf). ,9,
Step 2. We derive the polar equation for the orbit. Before doing so, however,
note the following result, whose proof is left to you as an exercise:
PROPOSITION 1.7 If x(r) has constant length (i.e., ||x(f)|| is constant for
all r), then x is perpendicular to its derivative dx/dt.
Continuing now with the main argument, note that the vector r(r) is denned
so that its magnitude is precisely the polar coordinate r of the planet's position.
Using equations (8) and (9), we find that
axe
GM
u x r u x
du
dt
= -GM
= GM
duX
U X | U X
dt )
du\
U X X u
dt J
= GM
= GM
(u-u)
du
dt
du
u- — | u
dt
du
1 Ou
dt
(see Exercise 27 of § 1 .4)
(by Proposition 1.7)
= - (GMu),
dt
since G and M are constant. On the other hand, we can "reverse" the product rule
to find that
d\
a x c = — x c
dt
dy dc
= — xc + vx— (since c is constant)
dt dt
= — (v X c).
dt
3.1 ; Parametrized Curves and Kepler's Laws
Figure 3.1 1 The angle 9 is
the angle between r and d.
Thus,
and, hence,
d d
a x c = — (GMu) = — (v x c),
dt dt
v x c = GMu + d,
(10)
where d is an arbitrary constant vector. Because both v x c and u lie in the xy-
plane, so must d.
Let us adjust coordinates, if necessary, so that d points in the i-direction (i.e.,
so that d = di for some d e R). This can be accomplished by rotating the whole
set-up about the z-axis, which does not lift anything lying in the xy-plane out of
that plane. Then the angle between r (and hence u) and d is the polar angle 6 as
shown in Figure 3.11.
By Theorem 3.3 of Chapter 1,
u-d= ||u|
Since c = ||c||,
2
Idll cos# = dcosi
(11)
c • c
= (r x v) • c
= r • (v x c)
= ru-(GMu + d)
Hence,
(Why? See formula (4) of §1.4.)
by equation (10).
GMr + rd cos6»
by equation (11). We can readily solve this equation for r to obtain
c2
r =
(12)
GM + d cosO
the polar equation for the planet's orbit.
Step 3. We now check that equation (12) really does define an ellipse by
converting to Cartesian coordinates. First, we'll rewrite the equation as
c2 (c2/GM)
GM + d cosO ~ l + (d/GM)cose'
and then let p = c2/ GM, e = d/ GM for convenience. (Note that p > 0.) Hence,
equation (12) becomes
r = ■ (13)
1 + e cos e
A little algebra provides the equivalent equation,
r = p — er cos6. (14)
Now r cose = x (x being the usual Cartesian coordinate), so that equation (14)
is equivalent to
r = p — ex.
To complete the conversion, we square both sides and find, by virtue of the fact
that r2 = x2 + y2,
x2 + y2 = p2
2pex + e2x2 .
198 Chapter 3 I Vector-Valued Functions
A little more algebra reveals that
(1- e2)x2 + 2pex + y2 = p2. (15)
Therefore, the curve described by the preceding equation is an ellipse if 0 <
\e\ < 1, a parabola if e = ±1, and a hyperbola if \e \ > 1 . Analytically, there is no
way to eliminate the last two possibilities. Indeed, "uncaptured" objects such as
comets or expendable deep space probes can have hyperbolic or parabolic orbits.
However, to have a closed orbit (so that the planet repeats its transit across the
sky), we are forced to conclude that the orbit must be elliptical.
More can be said about the elliptical orbit. Dividing equation (15) by 1 — e2
and completing the square in x, we have
\ 2 2 2
x +
1 -e2 (1 -e2)2'
This is equivalent to the rather awkward-looking equation
(x + pe/(l-e2)f y2
p2/(l - e2)2 p2/{\ - e2) y J
From equation (16), we see that the ellipse is centered at the point (— pe/(l — e2),
0), that its semimajor axis has length a = p/(l — e2), and that its semiminor axis
has length b = p/^/l — e2. The foci of the ellipse are at a distance
yV - b2
P2 P2 P\e\
(1 - e2)2 1 - e2
from the center. (See Figure 3.12.) Hence, we see that one focus must be at the
origin, the location of the sun. Our proof is, therefore, complete. ■
Fortunately, all the toil involved in proving the first law will pay off in proofs
of the second and third laws, which are considerably shorter. Again, we retain all
the notation we already introduced.
THEOREM 1 .8 (Kepler's second law) During equal intervals of time, aplanet
sweeps through equal areas with respect to the sun.
Figure 3.1 2 The ellipse of equation (16).
3.1 ; Parametrized Curves and Kepler's Laws
Po(r0, B0)
P(r, 6)
Figure 3.1 3 The shaded area A(9) is
given by \r2 dip.
PROOF Fix one point Pq on the planet's orbit. Then the area A swept between
Po and a second (moving) point P on the orbit is given by the polar area integral
A{6)
f6 1 2
L 2r
dip.
(See Figure 3.13.) Thus, we may reformulate Kepler's law to say that dA/dt is
constant. We establish this reformulation by relating dA/dt to a known constant,
namely, the vector c = r x v.
By the chain rule (in one variable),
dA _ dA d9
dt d6 dt '
By the fundamental theorem of calculus,
dA d f9 1 , , 1 r ,~
Hence,
dA _ 1 2 dO
dt ~ 2r dt'
(17)
Now, we relate c to dO/dt by means of equation (9). Therefore, we compute
u x du/dt in terms of 0. Recall that u = - r and r = r cos 0 i + r sin 6* j. Thus,
r
cos 0 i + sin 0 j
u
du
dt dt ' dt
Hence, it follows by direct calculation of the cross product that
dO dO
smtf — i + cosO — j.
c = r u x
du
dt
2d0
= r — k,
dt
so c = ||c|| = r2d0/dt, and equation (17) implies that
dA _ 1
~dt ~ 2C'
a constant.
(18)
THEOREM 1 .9 (Kepler's third law) If T is the length of time for one plane-
tary orbit, and a is the length of the semimajor axis of this orbit, then T2 = Ka3
for some constant K.
Chapter 3 | Vector-Valued Functions
3.1 Exercises
PROOF We focus on the total area enclosed by the elliptical orbit. The area of an
ellipse whose semimajor and semiminor axes have lengths a and b, respectively,
is nab. This area must also be that swept by the planet in the time interval [0, T].
Thus, we have
'T dA
— dt
dt
T 1
-cdt by equation ( 1 8)
nab = I
Jo
-L
Hence,
1
= 2CT-
T = 2-^, so r2=4"W. (19)
c c2
Now, b and c are related to a, so these quantities must be replaced before we are
done. In particular, from equation (16), b2 = p2/(l — e2), so
b2 = pa.
Also
P
c2
GM
(See equations (12) and (13).) With these substitutions, the result in (19) becomes
t2 = An2a2{pa) = /4^r2\ ^
pGM \Gm)
This last equation shows that T2 is proportional to a3, but it says even more:
The constant of proportionality 4n2/GM depends entirely on the mass of the
sun — the constant is the same for any planet that might revolve around the sun.
In Exercises 1-6, sketch the images of the following paths, us-
ing arrows to indicate the direction in which the parameter
increases:
. \x = It - 1
1. { , , , -1 < t < 1
[y = 3 - t
2. x(t) = e' i + <?"' j
_ \x = t cos t , ,
3. { . , -6n < t < 6jt
I y = t sin t ~ ~
. \x = 3 cost „ -
I y = 2 sin 2t ~ ~
5. x(r) = (f,3r2+l,0)
6. x(t) = (t, t2, t3)
Calculate the velocity, speed, and acceleration of the paths
given in Exercises 7—10.
7. x(r) = (3f-5)i + (2? + 7)j
8. x(f ) = 5 cos t i + 3 sin t j
9. x(r) = (t sin*, t cos t, t2)
10. x(?) = (e',e2',2e')
In Exercises 11—14, (a) use a computer to give a plot of the
given path x over the indicated interval for t; identify the di-
rection in which t increases, (b) Show that the path lies on the
given surface S.
^ 11. x(?) = (3cosjrf,4sin7r?,2?), -4 < t < 4; S is ellip-
x2 y2
tical cylinder \- - — = 1 .
9 16
^12. x(f) = (t cost, t shU, t), -20 < t < 20; S is cone
Z2 = x2 + y2.
^ 13. x(?) = (t sin2r, t cos2r, t2), -6 < t < 6; S is para-
boloid z = x2 + y2.
3.1 | Exercises 201
^ 14. x(r) = (2cosf, 2sinr, 3 sin8f), 0 < t < 2jt; S is cy-
linders2 + y2 = 4.
In Exercises 1 5-18, find an equation for the line tangent to the
given path at the indicated value for the parameter.
15. x(r) = te~> i + e3' j, t = 0
16. x(t) = 4cos?i— 3sinr j + 5tk, t = tt/3
17. x(t) = (t2,t\t5), t = 2
18. x(r) = (cos(e'), 3 - t2, t), t = 1
19. (a) Sketch the path x(f) = (t, f3 - It + 1).
(b) Calculate the line tangent to x when t = 2.
(c) Describe the image of x by an equation of the form
y = f(x) by eliminating t .
(d) Verify your answer in part (b) by recalculating the
tangent line, using your result in part (c).
Exercises 20-23 concern Roger Ramjet and his trajectory when
he is shot from a cannon as in Example 6 of this section.
20. Verify that Roger Ramjet's path in Example 6 is indeed
a parabola.
21 . Suppose that Roger is fired from the cannon with an
angle of inclination 6 of 60° and an initial speed Vq of
100 ft/sec. What is the maximum height Roger attains?
22. Suppose that Roger is fired from the cannon with an an-
gle of inclination 6 of 60° and that he hits the ground
1/2 mile from the cannon. What, then, was Roger's
initial speed?
23. If Roger is fired from the cannon with an initial speed of
250 ft/sec, what angle of inclination 6 should be used
so that Roger hits the ground 1 500 ft from the cannon?
24. Gertrude is aiming a Super Drencher water pistol at
Egbert, who is 1.6 m tall and is standing 5 m away.
Gertrude holds the water gun 1 m above ground at an
angle a of elevation. (See Figure 3.14.)
(a) If the water pistol fires with an initial speed of
7 m/sec and an elevation angle of 45°, does Egbert
get wet?
(b) If the water pistol fires with an initial speed of
8 m/sec, what possible angles of elevation will
cause Egbert to get wet? (Note: You will want to
use a computer algebra system or a graphics cal-
culator for this part.)
25. A malfunctioning rocket is traveling according to the
pathx(r) = (e2', 3f3 — 2t, t - j) in the hope of reach-
ing a repair station at the point (7e4, 35, 5). (Here
t represents time in minutes and spatial coordinates
are measured in miles.) Alt = 2, the rocket's engines
suddenly cease. Will the rocket coast into the repair
station?
26. Two billiard balls are moving on a (coordina-
tized) pool table according to the respective paths
x(f) = (t
!, j - 1J and y(f) = (t, 5 - r), where
t represents time measured in seconds.
(a) When and where do the balls collide?
(b) What is the angle formed by the paths of the balls
at the collision point?
27. Establish part 1 of Proposition 1.4 in this section: If x
and y are differentiable paths in R" , show that
dt
(x-y):
dx
dy
dt '
28. Establish part 2 of Proposition 1.4 in this section: If x
and y are differentiable paths in R3, show that
dt
dx dy
(xxy) = — xy + xx — .
dt at
29. Prove Proposition 1.7.
30. (a) Show that the path x(f) = (cos t, cos t sin t, sin2 t)
lies on a unit sphere.
(b) Verify that x(r) is always perpendicular to the ve-
locity vector v(f ).
(c) Use Proposition 1 .7 to show that if a differentiable
path lies on a sphere centered at the origin, then
its position vector is always perpendicular to its
velocity vector.
Figure 3.14 Figure for Exercise 24.
Chapter 3 | Vector-Valued Functions
31. Consider the path
x = (a + b cos cot ) cos t
■ y = (a + b cos cot) sin t ,
z = b sin cot
where a, b, and co are positive constants and a > b.
(a) Use a computer to plot this path when
i. a = 3, b = 1, and co = 15.
ii. a = 5, b = 1, and co = 15.
iii. a = 5, b = 1, and w = 25.
Comment on how the values of a, b, and <w affect
the shapes of the image curves.
(b) Show that the image curve lies on the torus
(vV + y2-a)2 + z2 = b2.
(A torus is the surface of a doughnut.)
32. For the path x(f ) = (e' cos t, e' sin f ), show that the an-
gle between x(t) and x'(t) remains constant. What is
the angle?
33. Consider the path x: R R2, x(f ) = (t2, t3 - t).
(a) Show that this path intersects itself, that is, that
there are numbers t\ and ti such that x(f i ) = x(?2).
(b) At the point where the path intersects itself, it
makes sense to say that the image curve has two
tangent lines. What is the angle between these tan-
gent lines?
34. Although the path x : [0, 2tt] — > R2, x(f ) =
(cosf, sinf) may be the most familiar way to give a
parametric description of a unit circle, in this problem
you will develop a different set of parametric equations
that gives the x- and v-coordinates of a point on the
circle in terms of rational functions of the parameter.
(This particular parametrization turns out to be useful
in the branch of mathematics known as number theory.)
To set things up, begin with the unit circle x1 +
y2 = 1 and consider all lines through the point (—1,0).
(See Figure 3.15.) Note that every line other than the
vertical line x = — 1 intersects the circle at a point
(x, y) other than (—1, 0). Let the parameter t be the
slope of the line joining (—1,0) and a point (x, y) on
the circle.
y
(-1,0)1
2" Slope?
X
Figure 3.15 Figure for Exercise 34.
(a) Give an equation for the line of slope t joining
(—1,0) and (x, y). (Your answer should involve
x, y, and t.)
(b) Use your answer in part (a) to write y in terms of
x and t. Then substitute this expression for y into
the equation for the unit circle. Solve the resulting
equations for x in terms of /. Your answer(s) for x
will give the points of intersection of the line and
the circle.
(c) Use your result in part (b) to give a set of paramet-
ric equations for points (x, y) on the unit circle.
(d) Does your parametrization in part (c) cover the
entire circle? Which, if any, points are missed?
35. Let x(f) be a path of class C1 that does not pass through
the origin in R3. If x(fo) is the point on the image of x
closest to the origin and x'(fo) / 0, show that the po-
sition vector x(?o) is orthogonal to the velocity vector
x'(fo).
3.2 Arclength and Differential Geometry
In this section, we continue our general study of parametrized curves in R3,
considering how to measure such geometric properties as length and curvature.
This can be done by defining three mutually perpendicular unit vectors that form
the so-called moving frame specially adapted to a path x. Our study takes us
briefly into the branch of mathematics called differential geometry, an area where
calculus and analysis are used to understand the geometry of curves, surfaces,
and certain higher-dimensional objects (called manifolds).
3.2 | Arclength and Differential Geometry
Length of a Path
For now, let x: [a, b] —> R3 be a C1 path in R3. Then we can approximate the
length L of x as follows: First, partition the interval [a, b] into n subintervals.
That is, choose numbers to, t\, ...,/„ such that a = to < t\ < ■ ■■ < t„ = b. If,
for i = 1, . . . , n, we let As, denote the distance between the points x(r,_i) and
x(ti) on the path, then
n
1=1
(See Figure 3.16.) We have x(f) = (x(t), y(t), z(t)), so that the distance formula
(i.e., the Pythagorean theorem) implies
Ast = J Ax2 + Ay,2 + Az2,
where Axj = x(t{) - x(ti-\), Ayt = y(u) - y(ti-\), and Az,- = z{U) - zft-i). It
is entirely reasonable to hope that the approximation in (1) improves as the Att 's
become closer to zero. Hence, we define the length L of x to be
n
L= lim Y^AXi2 + Ay,-2 + Az,-2. (2)
max Ar,-^0 ~—f
i=l
Now, we find a way to rewrite equation (2) as an integral. On each subinterval
fj], apply the mean value theorem (three times) to conclude the following:
1. There must be some number t* in [£,-_i, ?, ] such that
x(tt) - x(»,_i) = x'(ff)fe - fj_i);
that is, Ax, = x'{t*)Att.
2. There must be another number t** in [/, i , r,] such that
Ayi=y'(t**)Att.
3. There must be a third number t*** in [/,_!, f, ] such that
Az, =z'(t***)Ath
Therefore, with a little algebra, equation (2) becomes
L= lim V Jx'{tff + y'(t**)2 + z'(t***)2 Ath (3)
max Ar,^0 z — ■f V
( = 1
When the limit appearing in equation (3) is finite, it gives the value of the definite
integral
b
Jx>(t)2 + y>(t)2 + z'(t)2dt.
Note that the integrand is precisely ||x'(f)||, the speed of the path. (This makes
perfect sense, of course. Speed measures the rate of distance traveled per unit
time, so integrating the speed over the elapsed time interval should give the total
distance traveled.) Moreover, it's not hard to see how we should go about defining
the length of a path in R" for arbitrary n.
L
204 Chapter 3 | Vector- Valued Functions
DEFINITION 2.1 The length L(x) of a C1 path x: [a, b] -> R" is found by
integrating its speed:
= f llx'
J a
L(x) = / \\x'(t)\\dt.
Figure 3.17 AC1 path.
x(a)
Figure 3.1
x: [a, b]
x(b)
8 Apiecewise C1 path
R3.
EXAMPLE 1 To check our definition in a well-known situation, we compute
the length of the path
x: [0, 2n] — > R2, x(/) = (a cost, a sin t), a > 0.
We have
so
x'(r) = — a sinr i + a cos? j,
||x'(f)ll = V o2 sin2 1 + a2 cos2 f = a.
Thus, Definition 2.1 gives
L(x)
2,t
: dt = lit a.
Since the path traces a circle of radius a once, the length integral works out to be
the circumference of the circle, as it should. ♦
EXAMPLE 2 For the helix x(t) = (a cost, a sint, bt), 0 < t < 2ir, we have
x'(t) = —a sin t i + a cos t j + b k,
so that ||x'(0|| = Va2 + £2, and
L(x)
= / V<s2 + b2dt = + b2.
Jo
When b = 0, the helix reverts to a circle and the length integral agrees with the
previous example. ♦
Although we have defined the length integral only for C1 (or "smooth-
looking") paths, there is no problem with extending our definition to the piecewise
C1 case. By definition, a C1 path is one with a continuously varying velocity vec-
tor, and so it typically looks like the path in Figure 3. 1 7. A piecewise C 1 path is one
that may not be C1 but instead consists of finitely many C1 chunks. A continuous,
piecewise C1 path that is notC1 typically looks like the path in Figure 3. 18. Each
of the three portions of the path defined for (i) a < t < t\, (ii) t\ < t < h, and
(iii) t2 < t < b is of class C 1 , but the velocity, if nonzero, would be discontinuous
at t = t\ and t = t2 - To define the length of a piecewise C1 path, all we need do is
break up the path into its C1 pieces, calculate the length of each piece, and add to
get the total length. For the piecewise C 1 path shown in Figure 3.18, this means we
would take
rt\ pt2 rb
/ \\x'(t)\\dt+ \\x'(t)\\dt+ ||x'(f)||df
J a J t\ J t2
to be the length.
3.2 | Arclength and Differential Geometry
Warning Even if a path is continuous, the definite integral in Definition 2.1
may fail to exist. An example of such an unfortunate situation is furnished by the
pathx: [0, 1] R2,
1
t sin
x(t) = (t, y(t)), where y(t)
t sin - if t 0
0
iff = 0
Such a path is called nonrectifiable. It is a fact that any C1 path with endpoints
is rectifiable, which is why we made such a condition part of Definition 2.1.
The Arclength Parameter
The calculation of the length of a path is not only useful (and moderately inter-
esting) in itself, but it also provides a way for us to reparametrize the path with
a parameter that depends solely on the geometry of the curve traced by the path,
not on the way in which the curve is traced.
Let x be any C1 path and assume that the velocity x' is never zero. Fix a point
P0 on the path and let a be such that x{a) = Pq. We define a one-variable function
s of the given parameter t that measures the length of the path from Pq to any
other (moving) point P by
Figure 3.19 The arclength
reparametrization.
(See Figure 3.19. The Greek letter tau, t, is used purely as a dummy variable —
the standard convention is never to have the same variable appearing in both the
integrand and either of the limits of integration.) If t happens to be less than a,
then the value of s in formula (4) will be negative. This is nothing more than a
consequence of how the "base point" Po is chosen.
Here's how to get the new parameter: From formula (4) and from the funda-
mental theorem of calculus,
ds
dt
d
dt
J a
x\x)\\dx = \\x'(t)\\ = speed.
(5)
Since we have assumed that x'(t) ^ 0, it follows that ds/dt is nonzero. Hence,
ds/dt is always positive, so s is a strictly increasing function of t. Thus, s is,
in fact, an invertible function; that is, it is at least theoretically possible to solve
the equation s = s(t) for t in terms of s. If we imagine doing this, then we can
reparametrize the path x, using the arclength parameter s as independent variable.
EXAMPLE 3 For the helix x(t ) = (a cost, a sin t, bt), if we choose the "base
point" P0 to be x(0) = (a, 0, 0), then we have
s(t) = f \\x\r)\\dr= f ja2 + b2dx = ja2 + b2t,
Jo Jo
so that
s = y/a2 + b2t,
Chapter 3 | Vector-Valued Functions
or
s
Va2 + b2 '
(What the preceding tells us is that this reparametrization just rescales the time
variable.) Hence, we can rewrite the helical path as
/ / s \ ( s \ bs
x(s) = la cos . , a sin . , .
V \s/a2 + b2) \y/a2 + b2) Va2 + b2
EXAMPLE 4 The explicit determination of the arclength parameter for a given
parametrized path is a delicate matter. Consider the path
Thenx'(0 = (1, V2t, t2) and, if we take the base point to be x(0) = (0, 0, 0),then
s(t)= f y/l + 2x2 + x4dx
Jo
f' f' ?3
= / v/(1 + t2)2Jt = / (l + x2)dx = t + -.
Jo Jo 3
On the other hand the path y(t) = (?, f2, f3) is quite similar to x, yet it has
no readily calculable arclength parameter. In this case, y'(0 = (1, 2f, 3?2) and the
resulting integral for s(t) is
s(t)= f y/\+Ax2 + 9xAdx.
Jo
It can be shown that this integral has no "closed form" formula (i.e., a formula
that involves only finitely many algebraic and transcendental functions). ♦
The significance of the arclength parameter s is that it is an intrinsic param-
eter; it depends only on how the curve itself bends, not on how fast (or slowly)
the curve is traced. To see more precisely what this means, we resort to the chain
rule. Consider s as an intermediate variable and t as a final variable. Then we
have
ds
x(t) = x (s) — by the chain rule,
dt
= As)\\x'(t)\\ by (5).
Since x'(t) ^ 0, we can solve for x'(s) to find
x'(s)= — —. (6)
l|x'(OII
Therefore, x'(s) is precisely the normalization of the original velocity vector, and
so it is a unit vector. Hence, the reparametrized path x(s ) has unit speed, regardless
of the speed of the original path x(f). (This result makes good geometric sense,
too. If arclength, rather than time, is the parameter, then speed is measured in
units of "length per length," which necessarily must be one.)
The only unfortunate note to our story is that the integral in formula (4) is
usually impossible to compute exactly, thus making it impossible to compute s
as a simple function of t. (The case of the helix is a convenient and rather special
3.2 | Arclength and Differential Geometry
exception.) One generally prefers to work indirectly, letting the chain rule come
to the rescue. We shall see this indirect approach next.
The Unit Tangent Vector and Curvature
Let x: / c R — ■>• R3 be a C3 path and assume that x' is never zero.
DEFINITION 2.2 The unit tangent vector T of the path x is the normal-
ization of the velocity vector; that is,
v x'(0
T =
Ml l|x'(OII
We see from Definition 2.2 that the unit tangent vector is undefined when the
speed of the path is zero. Also note that, from equation (6), T is dx/ds, where s
is the arclength parameter. Geometrically, T is the tangent vector of unit length
that points in the direction of increasing arclength, as suggested by Figure 3.20.
EXAMPLE 5 For the helix x(f) = (a cost, a sin t, bt), we have
x'(f) — a sin? i + a cos t j + bk.
T(f) = — — = J .
I|x'(0ll Ja2 + b2
On the other hand if we parametrize the helix using arclength so that
bs
then
x(s) = a cos , ] , a sin .
\Vfl2 + W \y/a2 + b2) Va^Tb2
T(s) = x'O) = . = = sin i + cos
■s/a2 + b2 V V«2 + b2 ) y/a2 + b2 \-Ja2 + b2
b
+
Va2 + b2
This agrees (as it should) with the first expression for T, since s = -J a2 + b2 t,
as shown in Example 3. ♦
Using the unit tangent vector, we can define a quantity that measures how
much a path bends as we travel along it. To do so, note the following key facts:
PROPOSITION 2.3 Assume that the path x always has nonzero speed. Then
1. dT/dt is perpendicular to T for all t in / (the domain of the path x).
2. \\dT/dt || \t=to equals the angular rate of change (as t increases) of the direc-
tion of T when t = to.
PROOF (You can omit reading this proof for the moment if you are interested in
the main flow of ideas.) To prove part 1 , we have
T(0-T(0 = 1,
Chapter 3 | Vector-Valued Functions
since T is a unit vector. Hence,
T(r0 + At)
T(t0)
Figure 3.21 The vector triangle
used in the proof of
Proposition 2.3.
dt
(T • T) = 0,
because the derivative of a constant is zero. Also we have
d dT dT
— (T-T) = T 1 T,
dt dt dt
by the product rule (Proposition 1.4). Thus,
dT
2T = 0.
dt
Therefore, T is always perpendicular to dT/dt. (See Proposition 1.7.)
Now we prove part 2. Because T is a unit vector for all t, only its direction
can change as t increases. This angular rate of change of T is precisely
lim — ,
Af->0+ At
where AO comes from the vector triangle shown in Figure 3.21. To make the
argument technically simpler, we shall assume that AT ^ 0. We claim that
AO
lim
Ar^0+ IIATI
1.
(7)
Then, from equation (7),
AO AO ||AT||
lim — = lim
AI-S-0+ At Af^0+ II AT
At
= lim
AO
lim
IATII
a^o+ II AT II Af^o+ At
= 1 • lim
IATII
Ar^0+ At
Since At is assumed to be positive in the limit, we may conclude that
lim
AT
dT
— - = lim
dt
At Ar^0+
At
as desired.
To establish equation (7), the law of cosines applied to the vector triangle in
Figure 3.21 implies
|| AT||2 = ||T(f + Af)||2 + ||T(f)||2 - 2||T(t + At)|| ||T(t)|| cos AO
= 2 - 2 cos AO,
because T is always a unit vector. Thus,
AO
lim
lim
AO
Ar^o+ ||AT|| a^o+ 72 - 2 cos AG
AO
= lim -
A'^0+ ^2 • 2(sin2(A6>/2))
from the half-angle formula, and so
AO AO/2
lim = lim = 1,
Ar^o+ ||AT|| Ar^o+ sin(A<9/2)
from the well-known trigonometric limit (or from L'Hopital's rule).
3.2 j Arclength and Differential Geometry 209
Part 2 of Proposition 2.3 provides a precise way of measuring the bending of
a path.
DEFINITION 2.4 The curvature k of a path x in R3 is the angular rate of
change of the direction of T per unit change in distance along the path.
The reason for taking the rate of change of T per unit change in distance in the
definition of k is so that the curvature is an intrinsic quantity (which we certainly
want it to be). Figure 3.22 should help you develop some intuition about k.
Figure 3.22 In the left figure, k is not large, since the
path's unit tangent vector turns only a small amount per
unit change in distance along the path. In the right
figure, k is much larger, because T turns a great deal
relative to distance traveled.
Because ||dT/<2r|| measures the angular rate of change of the direction of T
per unit change in parameter (by part 2 of Proposition 2.3) and ds/dt is the rate
of change of distance per unit change in parameter, we see that
where the last equality holds by the chain rule. It is formula (8) that we will use
when making calculations.
EXAMPLE 6 For the circle x(?) = {a cos t, a sin t), 0 < t <2tt,
ds
x (?) = — a sin? i + a cost j, ||x (?)|| = — = a,
dt
so that
xYf)
T(?) = = — sin? i + cos? j.
I|x'(?)||
Hence,
\\dT/dt\\ 1 . 1
k = = -|| -cos?i-sm?j|| = -
ds/dt a a
Thus, we see that the curvature of a circle is always constant with value equal
to the reciprocal of the radius. Therefore, the smaller the circle, the greater the
curvature. (Draw a sketch to convince yourself.) ♦
210
Chapter 3 | Vector-Valued Functions
EXAMPLE 7 If a and b are constant vectors in R3 and the path
x(t) = a t + b
traces a line. We have
so
x'(0 = a,
ds
Hence,
dt
T(/) =
= a .
which is a constant vector. Thus, T'(f) = 0 and formula (8) implies immedi-
ately that k is zero, which agrees with the intuitive fact that a line doesn't
curve. ♦
EXAMPLE 8 Returning to our friend the helix
x(t) = (a cos t , a sin t , bt),
we have already seen that
ds
dt
= ja2 + b2 and T(f) =
-a sin / i + a cos / j + b k
Va2 + b2
Thus, formula (8) gives
1
sja2 + b2
-a cos / i — a sin? j
Va2 + b2
a2 + b2'
We see that the curvature of the helix is constant, just like the circle. In fact, as b
approaches zero, the helix degenerates to a circle, and the resulting curvature is
consistent with that of Example 6.
We can also compute the curvature from the parametrization given by arc-
length. The same helix is also described by
\(s) = a cos
Vfl2 + b2
, a sin
bs
vV + W Va2 + W
and we have
dx
T(s) =
ds
sin
+
Va2 + b2 \Ja2 + b2
b
+
cos
Vfl2 + b2 \y/a2 + b2
■s/a2 + b2
We can, therefore, compute
dT a
ds
cos
a2 + b2 \Ja2 + b2
and hence, from formula (8), that
dT
sin
+ b2 \s/a2 + b2
h
ds
a2 + b2
which checks.
3.2 | Arclength and Differential Geometry
The Moving Frame and Torsion
We now introduce a triple of mutually orthogonal unit vectors that "travel" with a
given path x: / —> R3, known as the moving frame of the path. (Note: In general,
the term "frame" means an ordered collection of mutually orthogonal unit vectors
in R".) These vectors should be thought of as a set of special vector "coordinate
axes" that move from point to point along the path.
To begin, assume that (i) x'(0 # 0 and (ii) x'(t) x x"(f) ^ 0 for all t in /.
(The first condition assures us that x never has zero speed and the second that x
is not a straight-line path.) Then the first vector of the moving frame is just the
unit tangent vector:
dx = x'(Q
ds ||x'(OII '
(Now you see why condition (i) is needed.) For a second vector orthogonal to T,
recall that part 1 of Proposition 2.3 says that dT/dt must be perpendicular to T.
Hence, we define
(That dl/dt is not zero follows from assumptions (i) and (ii).) The vector N is
called the principal normal vector of x. By the chain rule, N is also given by
dT/ds
N = . (10)
\\dT/ds\\
Since k = \\dT/ds\\ by formula (8), we also see that
At a given point P along the path, the vectors T and N (and also the vectors
x' and x") determine what is called the osculating plane of the path at P. (See
Figure 3.23.) This is the plane that "instantaneously" contains the path at P. (More
Osculating plane
Figure 3.23 The osculating plane of the path x at the
point P.
212 Chapter 3 | Vector- Valued Functions
precisely, it is the plane obtained by taking points P\ and P2 on the path near P
and finding the limiting position of the plane through P, Pi, and P% as Pi and
P2 approach P along x. The word "osculating" derives from the Latin osculare,
meaning "to kiss.")
Now that we have defined two orthogonal unit vectors T and N, we can
produce a third unit vector perpendicular to both:
B = T x N.
(12)
The vector B, called the binormal vector, is defined so that the ordered triple
(T, N, B) is a right-handed system. Thus, B is a unit vector since
IBM = II T II
sin - = 1 ■ 1 • 1 = 1.
2
EXAMPLE 9 For the helix x(/) = (a cos t, a sin t, bt), the moving frame vec-
tors are
T(f) =
-a sin t i + a cos t j + b k
■J a1 + b2
(as we have already seen),
T'(f) (—a cos t i — a sinf j)/V«2 + b2
fs(t) = — — = ^=^= = — cos 1 1 — sin t j ,
|T'(t)ll
a/Va2 + b2
and
B(0 = T x N =
i j k
-a sinf/Va2 + bz a cos?/V«2 + b2 b/-Ja2 + b2
— cos t — sin t 0
sinf i- , , Fi cosf j+ / i k-
\«Ja2 + b2 J \V<32 + b2 J Wfl2 + b2
Equation (11) says that the derivative of T (with respect to arclength) is a
scalar function (namely, the curvature) multiple of the principal normal N. This
is not surprising, since N is defined to be parallel to the derivative of T. A more
remarkable result (see the addendum at the end of this section) is that the derivative
of the binormal vector is also always parallel to the principal normal; that is,
dB
— = (scalar function) N.
ds
The standard convention is to write this scalar function with a negative sign, so
we have
dB
— = -tN. (13)
ds
The scalar function r thus defined is called the torsion of the path x. Roughly
speaking, the torsion measures how much the path twists out of the plane, how
3.2 | Arclength and Differential Geometry
"three-dimensional" x is. Note that, according to our conventions, the curvature
k is always nonnegative (why?), while r can be positive, negative, or zero.
EXAMPLE 10 Consider again the case of circular motion. Thus, let x(/) =
(a cost, a sin /). Then, as shown in Example 6,
T(0
x'(t)
|x'(OII
sin? i + cos? j, and k
dT
ds
Now we calculate
T'(0
N = = — cos M-sinn',
l|T'(0ll
B = TxN = k, a constant vector.
Hence, dB/ds = 0, so there is no torsion. This makes sense, since a circle does
not twist out of the plane. ♦
EXAMPLE 1 1 Let x(/) = (e' cos /, e' sin / , e'). We calculate T, N, and B and
identify the curvature and torsion of x.
To begin, we have
x'(t) e'(cos/ — sin/)i + e'(cos/ + sin/)j + e' k
||x'(0|| = Vfe<
1
71
((cos t — sin t) i + (cos t + sin t ) j + k) .
From this, we may compute
dT dT/dt ^(-(sin? + cos0i + (cosf - sinf)j)
ds ds/dt s/3e'
e-'
= — (— (sinf + cosf)i + (cos t — sinf)j),
so that the curvature is
dT
ds
sflt
3
Now we determine the remainder of the moving frame:
T'(0 1
N = -t: jr = —=(— (sinf + cos t)i + (cosf — sin/) j),
T'(0 V2
B = T x N = —((sinf - cos/)i - (sin/ + cos/)j + 2k).
V 6
Finally, to find the torsion, we calculate
dB _ dB/dt _ ^((cos/ + sin/)i + (sin/ -cos/)j)
ds ds/dt yfZe*
((cos f + sin /) i + (sin / — cos /) j)
3^2
-N,
so
T =
214 Chapter 3 | Vector-Valued Functions
Figure 3.24 Any vector in the
plane perpendicular to T can be
used for N.
EXAMPLE 12 If a and b are vectors in R3, then the straight-line path x(t) =
at + bhas, as we saw in Example 7, T = a/||a||. Thus, both dT/df anddT/ds are
identically zero. Hence, k = 0 (as shown in Example 7) and N cannot be defined
using formula (9). From geometric considerations, any unit vector perpendicular
to T can, in principle, be used for N. (See Figure 3.24.) If we choose one such
vector, then B can be calculated from formula (12). Since T, N, and B are all
constant, r must be zero. This is an example of a moving frame that is not
uniquely determined by the path x and serves to illustrate why the assumption
x' x x" / 0 was made. ♦
It is important to realize that the moving frame, curvature, and torsion are
quantities that are intrinsic to the curve traced by the path. That is, any parame-
trized path that traces the same curve (in the same direction) must necessarily
have the same T, N, B vector functions and the same curvature and torsion. This
is because all of these quantities can be denned entirely in terms of the intrinsic
arclength parameter s. (See Definition 2.2 and formulas (6), (8), (10), (1 1), (12),
and (13).)
Another important fact is that the curvature function k and the torsion function
t together determine all the geometric information regarding the shape of the
curve, except for the curve's particular position in space. To be more precise, we
have the following theorem, whose proof we omit:
THEOREM 2.5 Let s be the arclength parameter and suppose C\ and C2 are
two curves of class C3 in R3 . Assume that the corresponding curvature functions
k\ and K2 are strictly positive. Then if k\(s) = K2(s) and t\(s) = t2(s), the two
curves must be congruent (in the sense of high school geometry). In fact, given
any two continuous functions k and r, where k{s) > 0 for all s in the closed
interval [0, L], there is a unique curve parametrized by arclength on [0, L] (up to
position in space) whose curvature and torsion are k and r, respectively.
Tangential and Normal Components of Velocity and Accel-
eration; Other Curvature Formulas —
As we have seen, the moving frame provides us with an intrinsic set of vectors,
like coordinate axes, that are special to the particular curve traced by a path. In
contrast, the velocity and acceleration vectors of a path are definitely not intrinsic
quantities but depend on the particular parametrization chosen as well as on the
shape of the path. (The speed of a path is entirely independent of the geometry
of the curve traced.) We can get some feeling for the relationship between the
intrinsic notion of the moving frame and the extrinsic quantities of velocity and
acceleration by expressing the latter two vector functions in terms of the moving
frame vectors.
Thus, we begin with a C2 path x: / — > R3 having x' 7^ 0 and x' x x" ^ 0.
For notational convenience, let s denote ds/dt and s denote d2s/dt2. From
Definition 2.2, we know that T = v/||v|| and so, since the speed s = ds/dt = ||v||,
we have
v(f) = sT. (14)
3.2 | Arclength and Differential Geometry
Figure 3.25 Decomposition of
acceleration a into tangential and
normal components.
This formula says that the velocity is always parallel to the unit tangent vector,
something we know well. To obtain a similar result for acceleration, we can
differentiate (14) and apply the product rule:
a(0 = V(t)
dt
(sT) = ST + s
dT
dt '
(15)
Next, we express dT/dt in terms of the T, N, B frame. Formula (11) gives
the derivative of dT/ds in terms of N. The chain rule says that dT/ds =
(dT/dt)/(ds/dt). Thus, from formula (1 1), we have
dT dT
—— = s— = skN.
dt ds
Hence, we may rewrite equation (15) as
a(r) = j'T + k j2N.
(16)
Warning s = d2s/ dt2 is the derivative of the speed which is a scalar function.
The acceleration a is the derivative of velocity and so is a vector function.
Note that formula (16) shows that the acceleration has no component in the
direction of the binormal vector B. Therefore, both velocity and acceleration are
vectors that lie in the osculating plane of the path. (See Figure 3.25.)
At first glance, it may not appear to be especially easy to use formula (16)
to resolve acceleration into its tangential and normal components because of the
curvature term. However,
||a||2 = a • a = (j'T + k j2N) • (ST + /fj2N) = s2 + (ks2)2,
since T and N are perpendicular vectors. Consequently, we may calculate the
components as follows:
Tangential component of acceleration = atang = s.
Normal component of acceleration = an0rm = ks2 = J ||a||z — a
'tang-
EXAMPLE 1 3 Let x(Q = (t , It , t2). Then v(/) = i + 2j + 2/k and a(f ) = 2k.
We have s = ||v(f)ll = V5 + 4f2. Therefore,
At
V5 + At2 '
Since ||a|| = 2, we see that
I a II 2 - fitting
16/2 2^5
5 + 4f2 V5 + At 2 '
Formulas (14) and (16) enable us to find an alternative equation for the
curvature of the path. We simply calculate that
v x a = (iT) x (sT + /ci2N) = ss(T x T) + kP(T x N) = ks3B.
216 Chapter 3 I Vector- Valued Functions
Recalling that s = ||v||, we have, by taking magnitudes,
||v x a|| = k||v||3||B|| =K ||v||3,
since B is a unit vector. Thus,
||vxa||
(17)
K = z—.
IMP
This relatively simple formula expresses the curvature (an intrinsic quantity)
in terms of the nonintrinsic quantities of velocity and acceleration.
EXAMPLE 14 For the path \(t) = (2t3 + l,t4, t5), we have
and
You can check that
v(0 = 6th + 4f3j + 5r4k
a(0 = 12ri + I2t2\ + 20f3k.
|v|| = t2^25t4 + I6t2 + 36
and
||v x a|| = ||4f4(5f2i - I5tj + 6k)|| = 4fV25f4 + 225/2 + 36.
Therefore, formula (17) yields
|| v x a|| _ 4(25f4 + 225r2 + 36)1/2
* ~ ||v||3 ~ t2(25t4 + 16f2 + 36)3/2 '
which is certainly a more convenient way to determine curvature in this case. ♦
Summary
You have seen many formulas in this section, and, at first, it may seem difficult
to sort out the primary statements from the secondary results. We list the more
fundamental facts here:
For a path x: / — >• R3 :
Nonintrinsic quantities:
Velocity v(0 = x'(t).
Speed % = ||v(/)||.
dt
Acceleration a(t) = x"(?).
3.2 | Arclength and Differential Geometry
Arclength function: (See Figure 3.26.)
s(t) = / ||x'(t)||Jt (basepoint is Pq = x(a))
J a
Intrinsic quantities:
The moving frame:
Unit tangent vector T =
Principal normal vector N =
Binormal vector B =
Curvature k =
dB
Torsion r is denned so that
ds
dx
X
ds~ ~
II*
dT/ds
\\dT/ds\\
TxN.
dT
ds
-tN
At)
dT/dt
= \\dT/dt\
WdT/dtW
ds/dt
Additional formulas:
v(?) = s T (s is speed).
a(t) = ST + ks2 N (s is derivative of speed).
Ilv x all
Addendum: More About Torsion and the
Frenet-Serret Formulas
We now derive formula (13), the basis for the definition of the torsion of a curve.
That is, we show that the derivative of the binormal vector B (with respect to
arclength) is always parallel to the principal normal N (i.e., that dB/ds is a
scalar function times N). The two main ingredients in our derivation are part 1 of
Proposition 2.3 and the product rule.
We begin by noting that, since the ordered triple of vectors (T, N, B) forms a
frame for R3, any moving vector, including dB/ds, can be expressed as a linear
combination of these vectors; that is, we must have
dB
ds
a(s)T + Z?(s)N + c(«)B,
(18)
where a, b, and c are appropriate scalar- valued functions. (Because T, N, and
B are mutually perpendicular unit vectors, any (moving) vector w in R3 can be
decomposed into its components with respect to T, N, and B in much the same
way that it can be decomposed into i, j, and k components — see Figure 3.27.) To
find the particular values of the component functions a, b, and c, it turns out that
218 Chapter 3 | Vector-Valued Functions
we can solve for each function by applying appropriate dot products to equation
(18). Specifically,
dB
T = a(s)T • T + b(s)N • T + c(s)B ■ T
ds
= a(s) ■ 1 + b(s) • 0 + c(s) • 0
= a(s),
and, similarly,
dB dB
— -N = —.B = c(s).
ds as
From Proposition 1.7, dB/ds is perpendicular to B and, hence, c must be zero.
To find a, we use an ingenious trick with the product rule: Because T • B = 0, it
follows that d /ds(T • B) = 0. Now, by the product rule,
d dB dT
— (TB) = T 1 B.
ds ds ds
Consequently, (dB/ds) • T = —(dT/ds) • B. Thus,
, . dB ^ dT „
a(s) = T = B
ds ds
= -*rN-B by formula (11),
= 0,
and equation ( 1 8) reduces to
dB
= b(sys.
ds
No further reductions are possible, and we have proved that the derivative of B is
parallel to N. The torsion r can, therefore, be defined by r(s) = — b(s).
Formulas (1 1) and (13) gave us intrinsic expressions for dT/ds and dB/ds,
respectively. We can complete the set by finding an expression for dN/ds. The
method is the same as the one just used. Begin by writing
JN
— = o(j)T + MsyS + c(s)B, (19)
ds
where a, b, and c are suitable scalar functions. Taking the dot product of equation
(19) with, in turn, T, N, and B, yields the following:
JN
a(s) = — • T, b(s) = — • N, c(s) = — • B.
ds ds ds
The "product rule trick" used here then reveals that
i/N „ dT
a(s)= — .T = -N.—
ds ds
= — N-/cN by formula (11)
and
JN JB
c(s)= B = -N.—
ds
= — N • (-tN) by formula (13)
= T.
3.2 | Exercises 219
Moreover, we may differentiate the equation N • N = 1 to find
b(s)
dN dN
■ N = — N •
= -a:T + tB.
ds ds '
which implies that b(s) is zero. Hence, equation (19) becomes
dN
ds
The formulas for dT/ds, dN/ds, and dB/ds are usually taken together as
T'O) = kN
N'(i) = -kT + tB
B'O) = -tN
and are known as the Frenet-Serret formulas for a curve in space. They are so
named for Frederic-Jean Frenet and Joseph Alfred Serret, who published them
separately in 1852 and 1851, respectively The Frenet-Serret formulas give a
system of differential equations for a curve and are key to proving a result like
Theorem 2.5. They are often written in matrix form, in which case, they have an
especially appealing appearance, namely,
r "
0
K
0 "
~ T
N'
— K
0
X
N
B'
0
— T
0
B
3.2 Exercises
Calculate the length of each of the paths given in Exercises
1-6.
1.
x(f) =
(2t+ l,7-3*),-l <r <2
2.
x(f) =
f2i+ |(2f + l)3/2j, 0 < f < 4
3.
x(f) =
(cos 3?, sin3f, 2?3/2), 0 < t < 2
4.
x(f) =
7i + t] + t2k, 1 < t < 3
5.
x(f) =
(r3,3r2,6f), -1 <r <2
6.
x(f) =
(In (cos 0, cost, sin f), f < ? < f
7.
x(f) =
Qnt, t2/2, y/2t), 1 < t < 4
8.
x(f) =
(2t cosf,2f sinr,2V2f2), 0 < ? < 3
9.
The path x(f ) = (a cos3 1, a sin3 f), where a
tive constant, traces a curve known as an astroid or a
hypocycloid of four cusps. Sketch this curve and find
its total length. (Be careful when you do this.)
10. If / is a continuously differentiable function, show
how Definition 2.1 may be used to establish the
formula
L = / y/\+{f'(X)Ydx
J a
forthe length ofthe curve y = f(x) between (a, f(a))
and (b, f(b)).
1 1 . Use Exercise 10 or Definition 2.1 (or both) to calculate
the length of the line segment y = mx + b between
(xq, yo) and (xi, yi). Explain your result with an ap-
propriate sketch.
12. (a) Calculate the length of the line segment deter-
mined by the path
x(t) = (alt + bua2t + b2)
as f varies from to to t\ .
(b) Compare your result with that of Exercise 1 1 .
(c) Now calculate the length of the line segment deter-
mined by the path x(f ) = a t + b as t varies from
f0 to h .
13. This problem concerns the path x=|f — l|i+|r|j,
-2 < t < 2.
(a) Sketch this path.
(b) The path fails to be of class C1 but is piecewise
Cl . Explain.
(c) Calculate the length of the path.
14. Consider the path x(f) = (e~' cost, e~' sinf).
Chapter 3 | Vector-Valued Functions
(a) Argue that the path spirals toward the origin as
t -»- +00.
(b) Show that, for any a, the improper integral
r
J a
\W{t)\\dt
converges.
(c) Interpret what the result in part (b) says about the
path x.
1 5. Suppose that a curve is given in polar coordinates by
an equation of the form r = f(9), where / is of class
C1. Use Definition 2.1 to derive the formula
J a
(8)2 + f(6)2d6
for the length of the curve between the points (/(a), a)
and (/(/S), j3) (given in polar coordinates).
16. (a) Find the arclength parameter s = s(t) for the path
x(?) = eat cos bt i + eat sin bt j + eat k.
(b) Express the original parameter / in terms of s and,
thereby, reparametrize x in terms of s.
Determine the moving frame {T, N, B}, and compute the cur-
vature and torsion for the paths given in Exercises 1 7—20.
17. x(0 = 5cos3H + 6t j + 5 sin 3? k
18. x(f) = (sin? - t cos?)i + (cosf + t sinf)j + 2k,
t > 0
19. x(r) = (t, \(t + I)3/2, 1(1 - tf'2), -1 < t < 1
20. x(f) = (e2' sin/, e2' cosf, 1)
21. (a) Use formula (17) in this section to establish the
following well-known formula for the curvature
of a plane curve y = f(x):
[i + (/'(x))2F2'
(Assume that / is of class C2.)
(b) Use your result in (a) to find the curvature of
y = In (shut).
22. (a) Let x(s) = {x(s), y(s)) be a plane curve para-
metrized by arclength. Show that the curvature is
given by the formula
K = \x'y" -x"y'\.
path x and, separately, plot the curvature k as a function of t
over the indicated interval for t and value(s) of the constants.
^ 23. x(?) = (acos?, fcsinr), 0 < t < 2jt; a = 2,b=l
^24. x(r) = (2a(l + cos r) cos f, 2a(l + cos f) sin f), 0<
t < 2jt; a = 1
^ 25. x(r) = (2a cosf(l + cost) - a, 2a sinf(l + cosf)),
0 < t < 2ji; a = l
^^26. x(r) = (a sin nt, b sin mt), 0 < t < 2jt; a = 3,
b = 2, n = 4, m = 3
Find the tangential and normal components of acceleration for
the paths given in Exercises 27—32.
(b) Showthatx(s)=(i(l-.y2), ^cos"1 s - Wl - s2
is parametrized by arclength, and compute its
curvature.
In Exercises 23-26, (a) use a computer algebra system to cal-
culate the curvature k of the indicated path x and (b) plot the
>)
27. x(?) =
f2i + fj
28. x(?) =
(2t , e2')
29. x(r) =
(e' cos2f , e* sin 2?)
30. x(r) =
(4cos5f, 5 sin4f, 30
31. x(?) =
(t,t,r2)
32. x(?) =
1(1 — cos t )i + sinf j + ^ cos t k
33. (a) Show that the tangential and normal compo-
nents of acceleration atang and anorm satisfy the
equations
x x x
^tang
(b) Use these formulas to find the tangential and
normal components of acceleration for the path
x(f) = (?+2)i + f2j + 3fk
34. Use Exercise 33 to show that, for the plane curve
y = m,
f'(x)f"(x)
"tang — , )
yi+(/'«)2
\f"{*)\
^norm — ; ■
yi + (f'M)2
35. Establish the following formula for the torsion:
(v x a) • a'
II v x a[|2
36. Show that kx = — T' • B', where differentiation is with
respect to the arclength parameter s.
37. Show that if x is a path parametrized by arclength and
x' x x" ^ 0, then
K2T = (X'XX")-X"'.
38. Suppose x: / -> R3 is a path with x'(f) x x"(r) / Ofor
all t e /. The osculating plane to the path at t = to is
the plane containing x(/o) and determined by (i.e., par-
allel to) the tangent and normal vectors T(fo) and N(fo).
3.3 | Vector Fields: An Introduction
The rectifying plane at t = to is the plane contain-
ing x(fo) and determined by the tangent and binormal
vectors T(fo) and B(?o). Finally, the normal plane at
t = to is the plane containing x(fo) and determined by
the normal and binormal vectors N(?o) and B(fo). Note
that both the osculating and rectifying planes may be
considered to be tangent planes to the path at to since
they are both parallel to T(fo).
(a) Show that B(fo) is perpendicular to the osculating
plane at to, that N(?o) is perpendicular to the rec-
tifying plane at to, and that T(to) is perpendicular
to the normal plane at to-
(b) Calculate the equations for the osculating, rec-
tifying, and normal planes to the helix x(?) =
(a cos t, a sin?, bi) at any ?o- (Hint: To speed your
calculations, use the results of Example 9.)
39. Recall that the equation for a sphere of radius a > 0
and center xo may be written as ||x — xo[| = a. (See
Example 15 of §2. 1 .) Explain why the image of a path
x with the property that
(x(f) - x0) • (x(r) - x0) = a2
for all t must lie on a sphere of radius a.
40. Let x be a path with x' x j" / 0 and suppose that there
is a point xo that lies on every normal plane to x. Show
that the image of x lies on a sphere. (See Exercise 38
concerning normal planes to paths.)
41 . Use the result of Exercise 40 to show that x(? ) =
(cos2f, — sin2f, 2 cos/) lies on a sphere by showing
that (1,0,0) lies on every normal plane to x.
42. Use the result of Exercise 27 of § 1 .4 to show that
N x B = T and B x T = N.
As a result, we can arrange T, N, and B in a circle so
that they correspond, respectively, to the vectors i, j,
k appearing in Figure 1.54 and so that we may use a
mnemonic for identifying cross products that is similar
to the one described in Example 1 of § 1 .4.
Let x be a path of class C3, parametrized by arclength s,
with x' x x" / 0. We define the Darboux rotation vector (also
called the angular velocity vector) by
w = rT + x:B.
Note that w(.?o) is parallel to the rectifying plane to x(so). The
direction of the Darboux vector w gives the axis of the "screw-
like" motion of the path x and its length gives the angular
velocity of the motion. Exercises 43-45 concern the Darboux
vector.
43. Show that ||w[| = si k1 + r2. (Hint: The vectors T, N,
and B are pairwise orthogonal.)
44. (a) Use the Frenet-Serret formulas to establish the
Darboux formulas:
T' = w x T
N' = w x N
B' = w x B.
(b) Use the Darboux formulas to establish the Frenet-
Serret formulas. Hence the two sets of equations
are equivalent. (Hint: Use Exercise 42.)
45. Show that x is a helix if and only if w is a constant
vector. (Hint: Consider w' and use Theorem 2.5.)
/ / / /
Figure 3.28 The constant vector
field F(x) = i + j.
3.3 Vector Fields: An Introduction
We begin with a simple definition.
DEFINITION 3.1 A vector field on R" is a mapping
MC R" R".
We are concerned primarily with vector fields on R2 or R3 . fn such cases, we
adopt the point of view that a vector field assigns to each point x in X a vector
F(x) in R" , represented by an arrow whose tail is at the point x. This perspective
allows us to visualize vector fields in a reasonable way.
EXAMPLE 1 Suppose F: R2 -■>• R2 is defined by F(x) = a, where a is a con-
stant vector. Then F assigns a to each point of R2, and so we can picture F by
drawing the same vector (parallel translated of course) emanating from each point
in the plane, as suggested by Figure 3.28. ♦
Chapter 3 | Vector-Valued Functions
(x, y)
G(x, y)
(0,0)
0
(1,0)
-j
(0,1)
i
(1,1)
i- j
Figure 3.29 The vector field
G(x, y) = yi — xj of Example 2.
EXAMPLE 2 Let's depict G: R2 -> R2, G(x, y) = yi- xj. We can begin to do
this by calculating some specific values of G, as in the adjacent table. However,
it is difficult to get much of a feeling for G as a whole in this way. To understand
G somewhat better, we need to "play around" a bit. Note that
IGCoOII = llvi-xjl
= r
where r = xi + yj, the position vector of the point (x, y). From this observation,
it follows that G has constant length a on the circle x2 + y2 = a2. In addition,
we have
r • G(x, y) = (xi + yj) • (yi - xj) = 0.
Hence, G(x, y) is always perpendicular to the position vector of the point (x, y).
These facts, together with a table like the preceding one, make it possible to see
that G looks like Figure 3.29. ♦
Remark Sometimes a scalar- valued function f:X c R" — >• R is called a scalar
field. One thinks of a vector field on R" as attaching vector information (such
as wind velocity) to each point and a scalar field as attaching real number infor-
mation (such as temperature or pressure). We'll use the term "scalar field" only
occasionally, but we don't want to shock you when we do.
EXAMPLE 3 Let r = xi + yj + zk. The so-called inverse square vector field
in R3 is a function F: R3 — {0} — > R3 given by
F(x, y, z) =
where c is any (nonzero) constant. If the term "inverse square" seems inappropriate
to you, we'll try to convince you otherwise. Setu = r/||r|| sothatr = ||r||u. Then
F is given by
F(x, y, z) =
Figure 3.30 An inverse square
vector field.
(1)
Therefore, F is a vector field whose direction at the point P(x, y, z) ^ (0, 0, 0)
is parallel to the vector from the origin to P and whose magnitude is inversely
proportional to the square of the distance from the origin to P. Note that F points
away from the origin if c is positive and toward the origin if c is negative.
We have seen an example of an inverse square field in §3.1 — namely, the
Newtonian gravitational field between two bodies. If one of the bodies is at the
origin and the other at (x, y, z), then we have
GMm
F= ^u.
In this case, the proportionality constant c is —GMm, which is negative. This
means that the gravitational force is attractive (i.e., it points in the direction
that reduces the distance between the two bodies). Such a vector field is shown in
Figure 3 .3 0. An example of a repelling inverse square field is the electrostatic force
3.3 | Vector Fields: An Introduction
between two particles with like static charges (both positive or both negative).
This force is expressed by Coulomb's law,
F = z- u,
where r is the vector from particle 1 (at the origin) to particle 2, u = r/||r||,
q\ and qi are the respective charges (positive or negative) on the particles, and
k is a constant appropriate for the units being used. In mks units, distance is
measured in meters, charge in coulombs, force in newtons, so that k is equal to
8.9875 x 109 Nm2/C2. ♦
Gradient Fields and Potentials
Inverse square fields are interesting not only for their origin in basic physical
situations, but also because they are examples of gradient fields. A gradient field
on R" is a vector field F: X c R" — > R" such that F is the gradient of some
(differentiable) scalar- valued function /: X -> R. That is,
F(x) = V/(x)
at all x in X. The function / is called a (scalar) potential function for the vector
field F. To see what this means in the case of the inverse square field (1), we write
out the components of F explicitly:
xi + yj + zk \
||r||2 V*2 + y2 + z2J \jx2 + y2 + z2
since r = xi + yj + zk and u = r/||r||. That is,
(X' y' Z) ~ (x2 + y2 + z2)3/2 1 + (x2 + y2 + z2)3/2 J + (x2 + y2 + z2)3/2 '
We leave it to you to check that F(x, y, z) = V/(x, y, z), where /:R3 — {0} -> R
is given by
f(x, y,z)
c
y/x2 + y2 + z2 Ilrll
Remark In physics and engineering, a negative sign is often introduced in the
definition of a potential function (i.e., so that a potential function g for a vector
field F is one such that F = — Vg). The motivation behind such a convention is
that in physical applications, it is desirable to have the potential function rep-
resent potential energy in some sense. For example, in the case of the gravita-
tional field F = — ( G M/w / 1 1 r 1 1 2 )u, a physicist would take the potential function to
be —GMm/\\r\\, not +GMm/||r|| as we do. The advantage to the physicist in
doing so is that the physicist's potential function increases with increasing ||r||.
This corresponds to the notion that the greater the distance between two bodies,
the greater should be the stored gravitational potential energy.
From Theorem 6.4 of Chapter 2 we know that the gradient of any C1 scalar-
valued function /: X c R" ->• R is perpendicular to the level sets of /. Thus, if
F is a gradient vector field on R" , F(x) must be perpendicular to the level set of a
224 Chapter 3 I Vector- Valued Functions
Figure 3.31 A gradient vector field F = V/. Equipotential
lines are shown where / is constant.
potential function of F containing the point x. If / is such a potential function, the
level set {x | f(x) = c] is called an equipotential set (or equipotential surface if
n = 3, or equipotential line if n = 2) of the vector field F. (See Figure 3.31.)
You've seen examples of equipotential lines every time you've looked at a
weather map. Usually curves of constant barometric pressure (called isobars) or
of constant temperature (isotherms) are drawn. ( See Figure 3.32.) Perpendicular
to such equipotential lines are associated gradient vector fields that point in the
direction of most rapid increase of pressure or temperature.
Flow Lines of Vector Fields
When you draw a sketch of a vector field on R2 or R3, it is easy to imagine
that the arrows represent the velocity of some fluid moving through space as in
Figure 3.33. It's natural to let the arrows blend into complete curves. What you're
Figure 3.32 A weather map. (Weather graphics courtesy of Accuweather, Inc. 385
Science Park Road, State College, PA 16803. (814) 237-0309. © 2011. Used with
permission.)
Figure 3.33 A fluid moving
through space.
3.3 I Vector Fields: An Introduction
225
doing analytically is drawing paths whose velocity vectors coincide with those of
the vector field.
Figure 3.34 A flow line.
Figure 3.35 The vector field
F(x,>>,z) = 2i-3j + kof
Example 4.
Figure 3.36 Flow lines of
F(x, y) = —yi + x\ of Example 5.
DEFINITION 3.2 A flow line of a vector field F: X C R" -> R" is a differ-
entiable path x: / — > R" such that
x'(t) = F(x(0).
That is, the velocity vector of x at time ? is given by the value of the vector
field F at the point on x at time ?. (See Figure 3.34.)
EXAMPLE 4 We calculate the flow lines of the constant vector field
F(x,y,z) = 2i-3j + k.
A picture of this vector field (see Figure 3.35) makes it easy to believe that
the flow lines are straight-line paths. Indeed if x(?) = (x(t), y(t), z(t)) is a flow
line, then, by Definition 3.2, we must have
x'(?) = (x'(?), y'(t), z'{t)) = (2, -3, 1) = F(x(?)).
Equating components, we see
V(0 = 2
y'(t) = -3.
z'(t)= 1
These differential equations are readily solved by direct integration; we obtain
x(?) = 2? + xq
y{t) = -3? + y0 ,
z(t) = t + zo
where xq, yo, and zo are arbitrary constants. Hence, as expected we obtain para-
metric equations for a straight-line path through an arbitrary point (x0, yo, Zo)
with velocity vector (2,-3,1). ♦
EXAMPLE 5 Your intuition should lead you to suspect that a flow line of the
vector field F(x, y) = — yi + xj should be circular as showninFigure 3.36. Indeed
if x: [0, 2jt) — > R2 is given by x(t) = (a cos t , a sin t), where a is constant, then
x'(f) = — a sinf i + a cost j = F(a cos/, a sin?),
so such paths are indeed flow lines.
Finding all possible flow lines of F(x , y) = — yi + xj is a more involved task.
If x(/) = (x(f), y(t)) is a flow line, then, by Definition 3.2, we must have
At) = x'(t)i + y'(t)j = -y(0i + *(0j = F(x(0).
Equating components,
\x'{t) = -y(t)
y'(t) = x(t) ■
This is an example of a first-order system of differential equations. It turns out
that all solutions to this system are of the form
x(t) = (a cos t — b sin?, a sin? + b cos ?),
Chapter 3 | Vector-Valued Functions
where a and b are arbitrary constants. It's not difficult to see that such paths trace
circles when at least one of a or b is nonzero. ♦
In general, if F is a vector field on R", finding the flow lines of F is equivalent
to solving the first-order system of differential equations
x[(t) = FiCti(f), X2(t), X„(t))
x'2(t) = F2(xi(t), x2(t), x„(t))
. X'„(t) = Fn(Xi(t), X2(t), . . . , Xn(t))
for the functions X\(t), . . . , xn(t) that are the components of the flow line x. (The
function F, is just the ith component function of the vector field F.) Such a
problem takes us squarely into the realm of the theory of differential equations, a
fascinating subject, but not of primary concern at the moment.
3.3 Exercises
In Exercises 1-6, sketch the given vector fields on R2.
1.
F
= yi-X\
2.
F
= xi- yj
3.
F
= {-*,y)
4.
F
= (x,x2)
5.
F
= (x2,x)
6.
F
= (y\y)
In Exercises 7-12, sketch the given vector field on R3.
7. F = 3i + 2j + k
8. F = (y, -x,0)
9. F = (0, z, -y)
10. F = (y, -x,2)
11. F = (y,-x,z)
V x
12. F= - y = i — = j
V-*2 + y2 + z2 V-*2 + y2 + z2
7
+ ; = =k
s/x2 + y2 + z2
In Exercises 13-16, use a computer to plot the given vector
fields over the indicated ranges.
^ 13. F = (x - y, x + y); -1 < x < 1, -1 < y < 1
^14. F = (y3x, x2y); -2 < x < 2, -2 < y < 2
15. F = (x siny, y cosx); —2jt<x<2jt,
—2it < y < 2tc
^ 16. F = (cos(x - y), sin(x + y)); -2n < x < 2jt,
—2it < y < 2tc
In Exercises 17-19, verify that the path given is a flow line of
the indicated vector field. Justify the result geometrically with
an appropriate sketch.
17. x(t) = (sinf, cosr, 0), F = (y, -x, 0)
18. x(r) = (sinf, cos?, 2t), F = (y, -x, 2)
19. x(r) = (sinf, cosf, e2t), F = (y, -x, 2z)
In Exercises 20-22, calculate the flow line x(f ) of the given
vector field F that passes through the indicated point at the
specified value of t.
20. F(x,y) = -xi + yj; x(0) = (2, 1)
21. F(x,y) = (x2,y); x(l) = (l,e)
22. F(x,y,z) = 2i-3yj + Z3k; x(0) = (3,5,7)
23. Consider the vector field F = 3 i - 2 j + k.
(a) Show that F is a gradient field.
(b) Describe the equipotential surfaces of F in words
and with sketches.
24. Consider the vector field F = 2x i + 2y j — 3k.
(a) Show that F is a gradient field.
(b) Describe the equipotential surfaces of F in words
and with sketches.
25. If x is a flow line of a gradient vector field F = V/,
show that the function G(f ) = /(x(f )) is an increasing
function of f. (Hint: Show that G'(t) is always non-
negative.) Thus, we see that a particle traveling along
a flow line of the gradient field F = V/ will move
from lower to higher values of the potential function
/. That's why physicists define a potential function of
a gradient vector field F to be a function g such that
F = — Vg (i.e., so that particles traveling along flow
lines move from higher to lower values of g).
3.4 | Gradient, Divergence, Curl, and the Del Operator 227
LetF: X C R" — > R" be a continuous vector field. Let (a, b)be
an interval in "Rthat contains 0. (Think of {a, b)asa "timeinter-
val") A flow ofF is a differentiable function cp: X x (a, b) — >
R" ofn + 1 variables such that
dt
0(x, t) = F(0(x, f)); 0(x, 0) = x.
Intuitively, we think of 0(x, t) as the point at time t on the flow
line ofF that passes through x at time 0. (See Figure 3.37.)
Thus, the flow ofF is, in a sense, the collection of all flow lines
ofF. Exercises 26-31 concern flows of vector fields.
F(0(x, 0)
<Hx,f)/
Figure 3.37 The flow of the vector field F.
26. Verify that
0:R2 x R^ R2,
<j>(x, .v, t)
x + y t ,x~y -t
e -\ e ,
2 2
x + y , y — x
e' + : e~
2 2
is a flow of the vector field F(x, y) = (y, x).
27. Verify that
</>:R2 x R^ R2,
4>(x, y, t) = (y sinf + x cos t, y cos? — x sinf)
is a flow of the vector field F(x, y) = (y, —x).
28. Verify that
0:R3 x R^ R3,
4>(x, y, z, t) = (x cos2f — y sin 2;, y cos2f
+ x sin2f, ze~')
is a flow of the vector field F(x,y,z)
2x j — z k.
-2vi +
29. Show that if <j>: X x (a, fo) R" is a flow ofF, then,
for a fixed point xo in X, the map x: (a, b) — ► R" given
by x(f) = 0(xo, f) is a flow line ofF.
30. If 0 is a flow of the vector field F, explain why
0(0(x, r), s) = 0(x, s + r). (Hint: Relate the value of
the flow 4> at (x, t) to the flow line ofF through x. You
may assume the fact that the flow line of a continuous
vector field at a given point and time is determined
uniquely.)
31 . Derive the equation of first variation for a flow of a
vector field. That is, if F is a vector field of class C1
with flow <p of class C2, show that
dt
Dx0(x, 0 = DF(0(x, t))D%<p(x, t).
Here the expression "Dx0(x, t)" means to differentiate
(p with respect to the variables x\ , X2, . . . , x„, that is,
by holding t fixed.
3.4 Gradient, Divergence, Curl, and the Del
Operator
In this section, we consider certain types of differentiation operations on vector
and scalar fields. These operations are as follows:
1. The gradient, which turns a scalar field into a vector field.
2. The divergence, which turns a vector field into a scalar field.
3. The curl, which turns a vector field into another vector field. (Note: The curl
will be denned only for vector fields on R3.)
We begin by defining these operations from a purely computational point of view.
Gradually, we shall come to understand their geometric significance.
The Del Operator
The del operator, denoted V, is an odd creature. It leads a double life as both
differential operator and vector. In Cartesian coordinates on R3, del is defined by
Chapter 3 | Vector-Valued Functions
the curious expression
The "empty" partial derivatives are the components of a vector that awaits suitable
scalar and vector fields on which to act. Del operates on (i.e., transforms) fields
via "multiplication" of vectors, interpreted by using partial differentiation.
For example, if /: X c R3 — >• R is a differentiable function (scalar field),
the gradient of / may be considered to be the result of multiplying the vector V
by the scalar /, except that when we "multiply" each component of V by /, we
actually compute the appropriate partial derivative:
/ 3 3 3 \ df df df
Vf(x,y,z)= i— +j— +k- )f(x,y,z)= -M+/j + /k.
\ dx ay dz ) dx dy dz
The del operator can also be defined in R", for arbitrary n. If we take
x\, X2, ■ ■ ■ , xn to be coordinates for R", then del is simply
where e, ■ = (0, . . . , 1, . . . , 0), i = 1, . . . , n, is the standard basis vector for R".
The Divergence of a Vector Field
Whereas taking the gradient of a scalar field yields a vector field the process of
taking the divergence does just the opposite: It turns a vector field into a scalar
field.
DEFINITION 4.1 Let F: X C R" R" be a differentiable vector field.
Then the divergence of F, denoted div F or V • F (the latter read "del dot
F"), is the scalar field
dFi dF2 dFn
div F = V • F = — L + — L + ... + — 1,
3xi dxi dx„
where Cartesian coordinates for R" and F\ , . . . , F„ are the
component functions of F.
It is essential that Cartesian coordinates be used in the formula of Definition 4.1.
(Later in this section we shall see what div F looks like in cylindrical and spherical
coordinates for R3 .)
EXAMPLE 1 IfF = jr2vi + xZj+xyzk,then
3 3d
div F = — (x2y) + — (jcz) + — (xyz) = 2xy + 0 + xy = 3xy. ♦
dx dy dz
3.4 | Gradient, Divergence, Curl, and the Del Operator 229
The notation for the divergence involving the dot product and the del operator
is especially apt: If we write
F = Fid + F2e2 H h Fnen,
then,
/ 3 3 3 \ ,
V-F= ei— +e2— + --- + e„— ■ (Fid + F2e2 + ■ ■ ■ + F„e„)
\ dXl 0X2 oxnJ
_ dFi dF2 dFn
dx\ 9x2 3x„ '
where, once again, we interpret "multiplying" a function by a partial differential
operator as performing that partial differentiation on the given function.
Intuitively, the value of the divergence of a vector field at a particular point
gives a measure of the "net mass flow" or "flux density" of the vector field in
or out of that point. To understand what such a statement means, imagine that
the vector field F represents velocity of a fluid. If V • F is zero at a point, then
the rate at which fluid is flowing into that point is equal to the rate at which
fluid is flowing out. Positive divergence at a point signifies more fluid flowing out
than in, while negative divergence signifies just the opposite. We will make these
assertions more precise, even prove them, when we have some integral vector
calculus at our disposal. For now, however, we remark that a vector field F such
that V • F = 0 everywhere is called incompressible or solenoidal.
EXAMPLE 2 The vector field F = xi + yj has
V-F= A(je)+ ^(y) = 2.
dx ay
This vector field is shown in Figure 3.38. At any point in R2, the arrow whose
tail is at that point is longer than the arrow whose head is there. Hence, there is
greater flow away from each point than into it; that is, F is "diverging" at every
point. (Thus, we see the origin of the term "divergence.")
The vector field G = — xi — yj points in the direction opposite to the vector
field F of Figure 3.38 (see Figure 3.39), and it should be clear how G's divergence
of —2 is reflected in the diagram. ♦
EXAMPLE 3 The constant vector field F(x, y, z) = a shown in Figure 3.40
is incompressible. Intuitively, we can see that each point of R3 has an arrow
representing a with its tail at that point and another arrow, also representing a,
with its head there.
The vector field G = yi — xj has
v-g= A(v) + A(_x) = o.
ax ay
A sketch of G reveals that it looks like the velocity field of a rotating fluid, without
either a source or a sink. (See Figure 3.41 .) ♦
The Curl of a Vector Field
If the gradient is the result of performing "scalar multiplication" with the del
operator and a scalar field, and the divergence is the result of performing the
"dot product" of del with a vector field, then there seems to be only one simple
Chapter 3 | Vector-Valued Functions
Figure 3.40 The constant vector
field F = a.
Figure 3.41 The vector field
G = yi — x'] resembles the
velocity field of a rotating fluid.
differential operation left to be built from del. We call it the curl of a vector field
and define it as follows:
DEFINITION 4.2 Let F: X c R3 -> R3 be a differentiable vector field on
R3 only. The curl of F, denoted curl F or V x F (the latter read "del cross
F"), is the vector field
(3 3 3 \
V + V + V ) x + F^ + F3k>
dx ay dz )
i j k
d/dx d/dy d/dz
F\ F2 F3
^EL - i + ( ?H _ ^fl\ ■ + ( ^fl _ |
dy dz J \ dz dx J \ dx dy
There is no good reason to remember the formula for the components of the
curl — instead simply compute the cross product explicitly.
EXAMPLE 4 If F = x2yi - 2xzj + (x + y - z)k, then
V x F
i j k
d/dx d/dy d/dz
— 2xz x + y — z
x2y
3 3 \ ( 3 3
— (x + y-z)- ^~2xz)) 1 + \dz~(x2y) ~ 9^(X + y ~ Z-
3
9
= (1 + 2x)i - j - (x2 + 2z)k.
3.4 | Gradient, Divergence, Curl, and the Del Operator 231
Figure 3.42 A twig in a pond where water moves with velocity given by a vector field F. In the left figure, the twig does
not rotate as it travels, so curl F = 0. In the right figure, curl F / 0, since the twig rotates.
One would think that, with a name like "curl," V x F should measure how
much a vector field curls. Indeed the curl does measure, in a sense, the twisting
or circulation of a vector field but in a subtle way: Imagine that F represents the
velocity of a stream or lake. Drop a small twig in the lake and watch it travel.
The twig may perhaps be pushed by the current so that it travels in a large circle,
but the curl will not detect this. What curl F measures is how quickly and in what
orientation the twig itself rotates as it moves. (See Figure 3.42.) We prove this
assertion much later, when we know something about line and surface integrals.
For now, we simply point out some terminology: A vector field F is said to be
irrotational if V x F = 0 everywhere.
EXAMPLE 5 Let F = (3x2z + y2) i + 2xy j + (x3 - 2z) k. Then
V x F
l
d/dx
3x2z + y2
a
j
d/dy
2xy
k
d/dz
2z
dy
(x> - 2z)
^(2xy)W(^(3x2z + y2)
dx
(x3-2z)
dx
(2xy)
3
9y
(3*2 z + y2) I k
= (0 - 0)i + (3x2 - 3x2)\ + (2y - 2y)k = 0.
Thus, F is irrotational.
Two Vector-analytic Results
It turns out that the vector field F in Example 5 is also a gradient field. Indeed
F = V/, where f(x, y, z) = x3z + xy2 — z2. (We'll leave it to you to verify
this.) In fact, this is not mere coincidence but an illustration of a basic result
about scalar-valued functions and the del operator:
THEOREM 4.3 Let /: X C R3 -> R be of class C2. Then curl (grad/) = 0.
That is, gradient fields are irrotational.
Chapter 3 | Vector-Valued Functions
PROOF Using the del operator, we rewrite the conclusion as
V x (V/) = 0,
which might lead you to think that the proof involves nothing more than noting
that V/ is a "scalar" times V, hence, "parallel" to V, so thatthe cross productmust
be the zero vector. However, V is not an ordinary vector, and the multiplications
involved are not the usual ones. A real proof is needed.
Such a proof is not hard to produce: We need only start calculating V x (V/).
We have
df df df
dx dy dz
Therefore,
V x (V/) =
i j k
d/dx d/dy d/dz
df/dx df/dy df/dz
d2f
dydz dzdy
dzdy) \
d2f d2f
dzdx dxdz
j +
d2f d2f
dxdy dydx
Since / is of class C2, we know that the mixed second partials don't depend
on the order of differentiation. Hence, each component of V x (V/) is zero, as
desired. ■
There is another result concerning vector fields and the del operator that is
similar to Theorem 4.3:
THEOREM 4.4 Let F:XcR3^R3 be a vector field of class C2. Then
div (curl F) = 0. That is, curl F is an incompressible vector field.
The proof is left to you.
EXAMPLE 6 If F = (xz — e2x cos z) i - yz j + e2l(sin y + 2 sin z) k, then
3 3 3
V • F = — (xz — e2x cosz) + — (-yz) + — (e2x(smy + 2sinz))
dx dy dz
= z - 2e2x cos z - z + 2e2x cos z = 0
for all (ij,z)£R5. Hence, F is incompressible. We'll leave it to you to check
that F = V x G, where G(x, y, z) = e2x cos y i + e2x sin z j + xyz k, so that, in
view of Theorem 4.4 the incompressibility of F is not really a surprise. ♦
Other Coordinate Formulations (optional)
We have introduced the gradient, divergence, and curl by formulas in Cartesian
coordinates and have, at least briefly, discussed their geometric significance. Since
certain situations may necessitate the use of cylindrical or spherical coordinates,
we next list the formulas for the gradient, divergence, and curl in these coordinate
systems. Before we do, however, a remark about notation is in order. Recall that
in cylindrical coordinates, there are three unit vectors er, e$, and ez that point in
the directions of increasing r, 6, and z coordinates, respectively. Thus, a vector
3.4 | Gradient, Divergence, Curl, and the Del Operator 233
field F on R3 may be written as
F = Frer + Fgee + F-e..
In general, the component functions Fr, Fg, and Fz are each functions of the
three coordinates r, 9, and z; the subscripts serve only to indicate to which of
the vectors e, , eg, and e; that particular component function should be attached.
Similar comments apply to spherical coordinates, of course: There are three unit
vectors ep, e9, and eg, and any vector field F can be written as
F = Fpep + Fvev + Feeg.
THEOREM 4.5 Let /: X c R3 -> R and F: Y C R3
scalar and vector fields, respectively. Then
3/ 13/ , 3/
— er H ee H e,;
9r r 30 3z
divF =
curl F =
3 3F0 3
or 39 dz
R be differentiate
(3)
(4)
er rt0
3/3r 3/36*
Fr rFg
e
3/3z
(5)
PROOF We'll prove formula (4) only, since the argument should be sufficiently
clear so that it can be modified to give proofs of formulas (3) and (5). The idea is
simply to rewrite all rectangular symbols in terms of cylindrical ones.
From the equations in (8) of § 1 .7, we have
e,- = cos 6 i + sin 9 j
eg = — sin 9 i + cos 6 j .
e- = k
(6)
From the chain rule, we have the following relations between rectangular and
cylindrical differential operators:
3 3 3
— = cos 9 h sin 6 —
3r dx 3y
3 3 3
— = —rsm9 h r cos 9 — .
89 dx 3y
3 _ 3
dz dz
These relations can be solved algebraically for d/dx, d/dy, and 3/3z to yield
a
= cos
3
9
sin 9
3
37
dr
r
39
3
= sin 6
3
Vr +
cos 9
8
3v"
r
39
3
3
dz
~ d~Z
(7)
234 Chapter 3 I Vector- Valued Functions
Hence, we can use (6) and (7) to rewrite the expression for the divergence of a
vector field on R3 :
V-F
3 3 3
IT 1 + IT J + IT k ) * (F'"e'- + FsCs + F^
dx dy dz
1 cos
,1
dr
sin# 3
~rr~d~9
j sin (9
3 cos9 3
8r
r 3(9
dz
[(Fr cos 9 - Fe sin 9) i + (Fr sin <9 + Fe cos 0) j + Fz k].
(We used the equations in (7) to rewrite the partial operators d/dx, 3/3 y, and
3/3z appearing in del and the equations in (6) to replace the cylindrical basis
vectors er, eg, and ez by expressions involving i, j, and k.) Performing the dot
product and using the product rule yields
V-F
cost
3 sva9 3
i
dr r d9
3 cos# 3
+ |sin6> 1
dr r d9
(Fr cos 9 — Fg sin#)
(Fr sin 9 + Fe cos 0) H F-
dz
dFr 3
= cost/ [cos 9 h Fr — (cos#)
dr dr
cos v sin
3r
dFr
dr
sin9
r
sin
+ sm9 \ sm9 — - + Fr — (sm.9)
dr dr
dF0 3
+ sin# | cosf h Fg — (cos$)
dr dr
dFr 3
cos 9 h Fr — (cos 9)
3(9 3(9
r
cos 9
r
cos 9
sin 9-
dFe
d9
3
Fe—(sin9)
3(9 v
dFr d
sm9 — - + Fr— (sine)
36> 89
cos9-
d_F^
89
3
Fa — (cos 9)
d9
+
dz
After some additional algebra, we find that
3 F
V • F = (cos2 9 + sin2 9) — - +
dr
sin2 9 + cos2 (
\Fr
+
dFr
sin2 6> + cos2 9 \ dFe
36
dFz
IF
1 1 dFe 8F,
dr r r d9 dz
(
\
r h Fr H rF,
3r 3(9 3zV
as desired.
In spherical coordinates, the story for the gradient, divergence, and curl is
more complicated algebraically, although the ideas behind the proof are essentially
3.4 I Exercises 235
the same. We state the relevant results and leave to you the rather tedious task of
verifying them.
THEOREM 4.6 Let /: X C R3 -> R and F: Y C R3 -> R3 be differentiable
scalar and vector fields, respectively. Then the following formulas hold:
v/ =
1 df
e
p d<p
1
p sin<p
deee;
V-F =
1 3 ,
pL dp
%) +
1 3
p sin<p 3^>
(sin<p Fy) +
1
eP
p sin <p e#
V x F =
d/dp
d/dcp
3/36*
p2 sin <p
p sin Fg
1
3Fe
p sin<p dO
(8)
(9)
(10)
3.4 Exercises
Calculate the divergence of the vector fields given in Exer-
cises 1-6.
1. F
3. F
4. F
5. F
6. F
x2i + y2]
2. F
y2i + x2}
(x + y)i + (y + z)j + (jc + z)k
Z cos (er ) i + xy/z2 + 1 j + e2v sin 3jc k
Xjei + 2x|e2 H h «x2e„
xiei + 2xie2 + • • • + nx\e„
Find the curl of the vector fields given in Exercises 7-11.
7. F = x2 i — xey j + 2xyz k
8. F = xi + yj + zk
9. F = (x + yz)i + (y + xz)} + (z + xy)k
1 0. F = (cos yz — x)i + (cos xz — y)j + (cos xy — z)k
11. F = y2zi + exyzj+x2yk
12. (a) Consider again the vector field in Exercise 8 and
its curl. Sketch the vector field and use your pic-
ture to explain geometrically why the curl is as you
calculated.
(b) Use geometry to determine V x F, where F =
(xi + yj + zk)
y/x2 + y2 + z2 '
(c) For F as in part (b), verify your intuition by explic-
itly computing V x F.
13. Can you tell in what portions of R2, the vector fields
shown in Figures 3.43-3.46 have positive divergence?
Negative divergence?
\ \ \ \ i i t
\ \ \ \ i i i
w \ u I I
v x \ \ \ \ I
/ / I 1 I i \
/ / I I I I I
////Ml
1 1 t t t / /
III////
I / / t f / /
I / / / / S S
I \ \ \ \ \ N
I I \ \ \ \ \
\ M \ \ \ \
Figure 3.43 Vector field for Exercise 13(a).
s \ \ \ \
i i / / /
S S / / / / \ \ \ \ *s ^ ^
Figure 3.44 Vector field for Exercise 13(b).
Chapter 3 | Vector-Valued Functions
Figure 3.45 Vector field for Exercise 13(c).
////II I
////ill
///iii i
✓ ✓/////
X \ \ \ \ \ \
S \ \ \ \ \ |
X \ \ \ \ \ (
\\ \ \ \ u
M \ \ V \ s
\ \ \ \ \ \ \
\ V \ \ \ V V
\ S \ \ \ \ N
III////
I I I I / / /
III////
III////
Figure 3.46 Vector field for Exercise 13(d).
14. Check that if f(x, y,z) = x2 sin y + y2 cosz, then
V x (V/) = 0.
15. Check that if F(x, y, z) = xyzi — ez cos xj + xy2z3K
then
V • (V x F) = 0.
16. Prove Theorem 4.4.
In Exercises 1 7-20, let r = x i + y j + zkand let r denote ||r||.
Verify the following:
17. Vr" =nr"-2r
18. V(lnr)= ^
ri
19. V • (r"r) = (n + 3)r"
20. V x (rnr) = 0
In Exercises 21—25, establish the given identities. (You may
assume that any functions and vector fields are appropriately
differentiable.)
21. V-(F + G) = V-F + V-G
22. V x (F + G) = V x F + V x G
23. V.(/F) = /V.F + F.V/
24. V x (/F) = /V x F + V/ x F
25. V-(FxG) = G- VxF-F-VxG
26. Prove formulas (3) and (5) of Theorem 4.5.
27. Establish the formula for the gradient of a function in
spherical coordinates given in Theorem 4.6.
28. The Laplacian operator, denoted V2, is the second-
order partial differential operator defined by
, s2 s2 a2
V2 = 1 1 .
dx2 3y2 dz2
(a) Explain why it makes sense to think of V2 as V • V .
(b) Show that if / and g are functions of class C2 , then
V2(/g) = fV2g + gV2f + 2(V/ • Vg).
(c) Show that
V-(/Vg-gV/) = /V2g-gV2/.
29. Show that V • (/V/) = || V/||2 + /V2/.
30. Show that V x (V x F) = V(V • F) - V2F. (Here
V2F means to take the Laplacian of each component
function of F.)
Let X be an open set in R", F:XC R" R" a vector field
on X, and a € X. If y is any unit vector in R", we define the
directional derivative of 'F at a in the direction of v, denoted
DvF(a), by
1
DvF(a) = lim -(F(a + hv) - F(a)),
h^O h
provided that the limit exists. Exercises 31-34 involve direc-
tional derivatives of vector fields.
31. (a) In analogy with the directional derivative of a
scalar-valued function defined in §2.6, show that
DvF(a)
dt
F(a + 1 v)
[=0
(b) Use the result of part (a) and the chain rule to show
that, if F is differentiable at a, then
DvF(a) = DF(a)v,
where v is interpreted to be an n x 1 matrix. (Note
that this result makes it straightforward to calculate
directional derivatives of vector fields.)
32. Show that the directional derivative of a vector field
F is the vector whose components are the directional
Miscellaneous Exercises for Chapter 3 237
derivatives of the component functions F\, . . . , F„ of
F, that is, that
DvF(a) = (DyFii*), DvF2(a), DvFn(a)).
33. LetF = vzi + xzj +xyk.FindD(i_j+k)/V3F(3,2, 1).
(Hint: See Exercise 31.)
34. Let F = x i + y j + z k. Show that DvF(a) = v for any
point a € R3 and any unit vector veR3. More gener-
ally, if F = (x\, X2, ■ ■ ■ , xn), a = (fli, ai, . . . , an), and
v = (t>i, V2 vn), show that _DvF(a) = v.
True/False Exercises for Chapter 3
1 . If a path x remains a constant distance from the origin,
then the velocity of x is perpendicular to x.
2. If a path is parametrized by arclength, then its velocity
vector is constant.
3. If a path is parametrized by arclength, then its velocity
and acceleration are orthogonal.
' ■x(f)|[ = ||x'(f)[|.
4.
dt
d
5. — (x x y)
dtK 3
6. K
dT
dt
dy dx
xx h y x — .
dt J dt
7. |T|
dB
ds
8. The curvature k is always nonnegative.
9. The torsion r is always nonnegative.
10. N
dT
ds
11. If a path x has zero curvature, then its acceleration is
always parallel to its velocity.
12. If a path x has a constant binormal vector B, then r =0.
|a(r)[|2
14. grad / is a scalar field.
15. div F is a vector field.
16. curl F is a vector field.
1 7. grad(div F) is a vector field.
18. div(curl(grad /)) is a vector field.
1 9. grad/ x div F is a vector field.
20. The path x(f) = (2 cos t, 4 sinr, t) is a flow line of the
y
vector field F(x ,y, z) = — — i + 2x j + z k.
21. The path x(?) = (e' cosf , e'(cos t + sinf), e' sin?) is a
flow line of the vector field F(jc, y, z) = (x — z)i +
2jc j + y k.
22. The vector field F = 2xy cos z i — y2 cos z j + exy k is
incompressible.
23. The vector field F = 2xy cos z i — y2 cos z j + exy k is
irrotational.
24. V x (V/) = 0 for all functions /: R3 — » R.
25. If V • F = 0 and V x F = 0, then F = 0.
26. V • (F x G) = F • (V x G) + G • (V x F).
27. If F = curl G, then F is solenoidal.
28. The vector field F = 2x sin y cos z i + x2 cos y cos z
j + x2 sin y sin z k is the gradient of a function / of
class C2.
29. There is a vector field F of class C2 on R3 such that
V x F = x cos2 y i + 3y j — xyz2 k.
30. If F and G are gradient fields, then F x G is incom-
pressible.
Miscellaneous Exercises for Chapter 3
1 . Figure 3.47 shows the plots of six paths x in the plane.
Match each parametric description with the correct
graph.
(a) x(/) = (s'mlt, sin 3i)
(b) x(f) = (t + sin5f, t2 + cos 6?)
(c) x(r) = (f2 + l,f3-?)
(d) x(f) = (2? + sin4f, t - sin5f)
(e) x(t) = {t -t2,t3 -t)
(f) x(t) = (sin(? + sin 30, cosf)
2. Figure 3.48 shows the plots of six paths x in R3 . Match
each parametric description with the correct graph.
(a) x(f) = (t + cos3f, t2 + sin5f, sin At)
(b) x(f) = (2 cos3 1, 3 sin3 1, cos2f)
(c) x(0 = (15 cos /, 23 sin t , At)
(d) x(0 = (cos3r, cos5f, sin4f)
(e) x(f) = (2f cosf, 2f sin?, 4f)
(f) x(f) = (r2 + l,?3-f,?4-r2)
238 Chapter 3 | Vector-Valued Functions
Miscellaneous Exercises for Chapter 3 239
3. Suppose that x is a C2 path with nonzero velocity. Show
that x has constant speed if and only if its velocity and
acceleration vectors are always perpendicular to one
another.
4. You are at Vertigo Amusement Park riding the new
Vector roller coaster. The path of your car is given by
x(?) = (x(t), y{t)), where
x(f)
t/60 1Tt t/60 ■ Jtt
e ' cos — ,e ' sm — ,
30 30
2r(10 - t)(t - 90)2
80 +
106
where / = 0 corresponds to the beginning of your
three-minute ride, measured in seconds, and spatial
dimensions are measured in feet. It is a calm day, but
after 90 sec of your ride your glasses suddenly fly off
your face.
(a) Neglecting the effect of gravity, where will your
glasses be 2 sec later?
(b) What if gravity is taken into account?
5. Show that the curve traced parametrically by
1
x(f) = ^cos(f - 1), t - 1,
istangenttothesurface.ii;3 + y3 + z3
t = 1.
xyz = 0 when
6. Gregor, the cockroach, is on the edge of a Ferris wheel
that is rotating at a rate of 2 rev/min (counterclock-
wise as you observe him). Gregor is crawling along
a spoke toward the center of the wheel at a rate of
3 in/min.
(a) Using polar coordinates with the center of the
wheel as origin, assume that Gregor starts (at time
t = 0) at the point r = 20 ft, 6 = 0. Give paramet-
ric equations for Gregor's polar coordinates r and
6 at time t (in minutes).
(b) Give parametric equations for Gregor's Cartesian
coordinates at time t.
(c) Determine the distance Gregor has traveled once
he reaches the center of the wheel. Express your
answer as an integral and evaluate it numerically.
If you have used a drawing program on a computer, you have
probably worked with a curve known as a Bezier curve.1 Such
a curve is defined parametrically by using several control
points in the plane to shape the curve. In Exercises 7-12,
we discuss various aspects of quadratic Bezier curves. These
curves are defined by using three fixed control points {x\ , y\),
(x2, yi), and(xi, yi) and a nonnegative constant w. The Bezier
curve defined by this information is given by x: [0, 1] — > R2,
X(t):
y(t) :
(1 - tfx\ + 2wt(l - t)x2 + t2X}
(\-tf + 2wt(\-t) + t2
(1 -tfyi +2wtQ_ -r)y2 + f2y3
(1 -t)2 + 2wt{\ -t) + t2
0 < t < 1.
(1)
^ 7. Let the control points be (1,0), (0, 1), and (1, 1).
Use a computer to graph the Bezier curve for w =
0, 1/2, 1, 2, 5. What happens as w increases?
8. Repeat Exercise 7 for the control points (—1, — 1),
(1,3), and (4, 1).
9. (a) Show that the Bezier curve given by the paramet-
ric equations in ( 1 ) has (x\ , y\ ) as initial point and
(x3, V3) as terminal point.
(b) Show that x(i) lies on the line segment joining
(*2, yi) to the midpoint of the line segment joining
(x\,y\) to (x3, y3).
10. In general the control points (xi,y{), (x2,y2), and
(*3 , )?3 ) will form a triangle, known as the control poly-
gon for the curve. Assume in this problem that w > 0.
By calculating x'(0) and x'(l), show that the tangent
lines to the curve at x(0) and x(l) intersect at (X2, V2).
Hence, the control triangle has two of its sides tangent
to the curve.
11. In this problem, you will establish the geometric sig-
nificance of the constant w appearing in the equations
in(l).
(a) Calculate the distance a between x( j ) and (x2 , ^2).
(b) Calculate the distance b between x(^) and the
midpoint of the line segment joining (xi, y\) and
C*3, yj)-
(c) Show that w = b/a. By part (b) of Exercise 9,
x(|) divides the line segment joining (X2, 3^) to
the midpoint of the line segment joining (x\, y\)
to (*3, V3) into two pieces, and w represents the
ratio of the lengths of the two pieces.
12. Determine the Bezier parametrization for the portion
of the parabola y = x2 between the points (—2, 4) and
(2, 4) as follows:
(a) Two of the three control points must be (—2, 4) and
(2, 4). Find the third control point using the result
of Exercise 10.
(b) Using part (a) and Exercise 9, we must have that
x(i) lies on the y-axis and, hence, at the point
2 P. Bezier was an automobile design engineer for Renault. See D. Cox, J. Little, and D. O'Shea, Ide-
als, Varieties, and Algorithms: An Introduction to Computational Algebraic Geometry and Commutative
Algebra, 3rd ed. (Springer- Verlag, New York, 2007), pp. 28-29. Exercises 7-1 1 adapted with permission.
240 Chapter 3 | Vector- Valued Functions
(0, 0). Use the result of Exercise 11 to determine
the constant w.
(c) Now write the Bezier parametrization. You should
be able to check that your answer is correct.
1 3. Let x: (0, 7t) -*■ R2 be the path given by
x(?) = (sin?, cos? + In tan |) ,
where ? is the angle that the y-axis makes with the
vector x(?). The image of x is called the tractrix. (See
Figure 3.49.)
(a) Show that x has nonzero speed except when ? =
n/2.
(b) Show that the length of the segment of the tangent
to the tractrix between the point of tangency and
the y-axis is always equal to 1. This means that
the image curve has the following description: Let
a horse pull a heavy load by a rope of length 1 .
of class C2. Use equation (17) in §3.2 to derive the
curvature formula
Figure 3.49 The tractrix
of Exercise 13.
Suppose that the horse initially is at (0, 0), the load
at (1, 0), and let the horse walk along the v-axis.
The load follows the image of the tractrix.
14. Another way to parametrize the tractrix path given in
Exercise 13 is
y: (-oo, O)-*- R2,
where y(r)
O-P
dp
(a) Show that y satisfies the property described in part
(b) of Exercise 13.
(b) In fact, y is actually a reparametrization of part of
the path x of Exercise 13. Without proving this fact
in detail, indicate what portion of the image of x
the image of y covers.
15. Suppose that a plane curve is given in polar coordi-
nates by the equation r = f{6), where / is a function
K(9)
' + 2r':
(r2 _|_ r/2)3/2
(Hint: First give parametric equations for the curve in
Cartesian coordinates using 9 as the parameter.)
16. Use the result of Exercise 15 to calculate the curvature
of the lemniscate r2 = cos 26.
Let x: / — > R2 be a path of class C2 that is not a straight line
and such that x'(?) ^ 0. Choose some to € I and let
y(?) = x(?) - s(f)T(f),
where s(t) = f' ||x'(r)|| dr is the arclength function and T is
the unit tangent vector. The path y: / —> R2 is called the invo-
lute of x. Exercises 1 7—19 concern involutes of paths.
17. (a) Calculate the involute of the circular path of radius
a, that is, x(?) = (a cos ?, a sin ?). (Take ?o to be 0.)
(b) Let a = 1 and use a computer to graph the path x
and the involute path y on the same set of axes.
1 8. Show that the unit tangent vector to the involute at ?
is the opposite of the unit normal vector N(?) to the
original path x. (Hint: Use the Frenet-Serret formulas
and the fact that a plane curve has torsion equal to zero
everywhere.)
19. Show that the involute y of the path x is formed by
unwinding a taut string that has been wrapped around
x as follows:
(a) Show that the distance in R2 between a point x(?)
on the original path and the corresponding point
y(?) on the involute is equal to the distance traveled
from x(?o) to x(?) along the underlying curve of x.
(b) Show that the distance between a point x(?) on the
path and the corresponding point y(?) on the in-
volute is equal to the distance from x(?) to y(?)
measured along the tangent emanating from x(?).
Then finish the argument.
Let x: / — > R2 be a path of class C2 that is not a straight line
and such that x'(?) / 0. Let
e(?) = x(?) + -N(?).
K
This is the path traced by the center of the osculating circle of
the path x. The quantity p = \/k is the radius of the osculat-
ing circle and is called the radius of curvature of the path x.
The path e is called the evolute of the path x. Exercises 20—25
involve evolutes of paths.
20. Letx(?) = (?, ?2)beaparabolicpath.(SeeFigure3.50.)
(a) Find the unit tangent vector T, the unit normal
vector N, and the curvature k as functions of ?.
(b) Calculate the evolute of x.
Miscellaneous Exercises for Chapter 3 241
(c) Use a computer to plot x(f) and e(?) on the same
set of axes.
y
Figure 3.50 The parabola and its
osculating circle at a point. The centers
of the osculating circles at all points of
the parabola trace the evolute of the
parabola as described in Exercise 20.
21 . Show that the evolute of a circular path is a point.
22. (a) Use a computer algebra system to calculate the for-
mula for the evolute of the elliptical path x(f) =
(a cost, b sinr).
(b) Use a computer to plot x(f ) and the evolute e(f ) on
the same set of axes for various values of the con-
stants a and b. What happens to the evolute when
a becomes close in value to bl
23. Use a computer algebra system to calculate the formula
for the evolute of the cycloid x(f ) = (at — a sin t, a —
a cost). What do you find?
24. Use a computer algebra system to calculate the formula
for the evolute of the cardioid x(f) = (2a cosf(l +
a cos t), 2a sin t (1 + a cos t )).
25. Assuming ic'(t) ^ 0, show that the unit tangent vector
to the evolute e(f ) is parallel to the unit normal vector
N(f ) to the original path x(f).
26. Suppose that a C1 path x(f ) is such that both its veloc-
ity and acceleration are unit vectors for all t . Show that
k = 1 for all t.
27. Consider the plane curve parametrized by
x(s) = / cos g(t)dt, y(s) = I sin g(t)dt,
Jo Jo
where g is a differentiable function.
(a) Show that the parameter s is the arclength param-
eter.
(b) Calculate the curvature k(s).
(c) Use part (b) to explain how you can create a
parametrized plane curve with any specified con-
tinuous, nonnegative curvature function k(s).
(d) Give a set of parametric equations for a curve
whose curvature k(s) = \s\. (Your answer should
involve integrals.)
(e) Use a computer to graph the curve you found in
part (d), known as a clothoid or a spiral of Cornu.
(Note: The integrals involved are known as Fres-
nel integrals and arise in the study of optics. You
must evaluate these integrals numerically in order
to graph the curve.)
28. Suppose that x is a C3 path in R3 with torsion r always
equal to 0.
(a) Explain why x must have a constant binormal vec-
tor (i.e., one whose direction must remain fixed for
all f).
(b) Suppose we have chosen coordinates so that x(0) =
0 and that v(0) and a(0) lie in the .ty-plane (i.e.,
have no k-component). Then what must the binor-
mal vector B be?
(c) Using the coordinate assumptions in part (b), show
that x(f) must lie in the jc;y-plane for all t . (Hint:
Begin by explaining why v(f) • k = a(r) • k = Ofor
all t. Then show that if
x(t) = x(t)i + y(t)j + z(t)k,
we must have z(t ) = 0 for all t .)
(d) Now explain how we may conclude that curves
with zero torsion must lie in a plane.
29. Suppose that x is a C3 path in R3, parametrized by arc-
length, with k / 0. Suppose that the image of x lies in
the xy -plane.
(a) Explain why x must have a constant binormal
vector.
(b) Show that the torsion r must always be zero.
Note that there is really nothing special about the im-
age of x lying in the Jfj-plane, so that this exercise,
combined with the results of Exercise 28, shows that
the image of x is a plane curve if and only if r is always
zero and if and only if B is a constant vector.
30. In Example 7 of §3.2 we saw that if x is a straight-line
path, then x has zero curvature. Demonstrate the con-
verse; that is, if x is a C2 path parametrized by arc-
length s and has zero curvature for all s, then x traces
a straight line.
31 . A large piece of cylindrical metal pipe is to be manu-
factured to include a strake, which is a spiraling strip
of metal that offers structural support for the pipe. (See
Figure 3.51.) The pieces of the strake are to be made
from flat pieces of flexible metal whose curved sides
are arcs of circles as shown in Figure 3.52. Assume that
242 Chapter 3 | Vector- Valued Functions
the pipe has a radius of a ft and that the strake makes
one complete revolution around the pipe every h ft.3
Figure 3.51 A cylindrical Figure 3.52 A section of
pipe with strake attached. the strake. (See Exercise 3 1 .)
(a) In terms of a and h, what should the inner radius r
be so that the strake will fit snugly against the pipe?
(b) Suppose a = 3 ft and h = 25 ft. What is r?
Suppose that x: / — > R3 is a path of class C3 parametrized by
arclength. Then the unit tangent vector T(s) defines a vector-
valued function T: / —> R3 that may also be considered to be a
path (although not necessarily one parametrized by arclength,
nor necessarily one with nonvanishing velocity). Since T is a
unit vector, the image of the path T must lie on a sphere of
radius 1 centered at the origin. This image curve is called the
tangent spherical image of x. Likewise, we may consider the
functions defined by the normal and binormal vectors N and B
to give paths called, respectively, the normal spherical image
and binormal spherical image of x. Exercises 32-35 concern
these notions.
32. Find the tangent spherical image, normal spherical
image, and binormal spherical image of the circular
helix x(/) = (a cosf , a sinf , bt). (Note: The path x is
not parametrized by arclength.)
33. Suppose that x is parametrized by arclength. Show
that x is a straight-line path if and only if its tangent
spherical image is a constant path. (See Example 7 of
§3.2 and Exercise 30.)
34. Suppose that x is parametrized by arclength. Show that
the image of x lies in a plane if and only if its binormal
spherical image is constant. (See Exercises 28 and 29.)
35. Suppose that x is parametrized by arclength. Show
that the normal spherical image of x can never be
constant.
36. In this problem, we will find expressions for velocity
and acceleration in cylindrical coordinates. We begin
with the expression
x(t) = x(t)i + y(t)j + z(t)k
for the path in Cartesian coordinates.
(a) Recall that the standard basis vectors for cylindri-
cal coordinates are
e,- = cosf? i + s'md j,
ee = — s'md i + cos# j,
e, = k.
Use the facts that x = r cos 0 and y = r sin 0 to
show that we may write x(r ) as
x(t) = r(t)er+z(t)ez.
(b) Use the definitions of er, ee, and ez just given and
the chain rule to find der/dt, deg/dt, and de./dt
in terms of er , eg , and ez .
(c) Now use the product rule to give expressions for v
and a in terms of the standard basis for cylindrical
coordinates.
37. Suppose that the path
x(f) = (sin2f, Vlcoslt, sin2f — 2)
describes the position of the Starship Inertia at time t .
(a) Lt. Commander Agnes notices that the ship is trac-
ing a closed loop. What is the length of this loop?
(b) Ensign Egbert reports that the Inertia's path is
actually a flow line of the Martian vector field
F(jc, y, z) = yi — 2x\ + yk, but he omitted a con-
stant factor when he entered this information in
his log. Help him set things right by finding the
correct vector field.
38. Suppose that the temperature at points inside a room is
given by a differentiable function T(x, y, z). Livinia,
the housefly (who is recovering from a head cold), is in
the room and desires to warm up as rapidly as possible.
(a) Show that Livinia 's path x(?) must be a flow line
of kVT, where & is a positive constant.
(b) If T(x, y, z) = x2 — 2y2 + 3z2 and Livinia is ini-
tially at the point (2,3,-1), describe her path
explicitly.
39. Let F = u(x, y)i — v(x, y)j be an incompressible,
irrotational vector field of class C2.
(a) Show that the functions u and v (which deter-
mine the component functions of F) satisfy the
Cauchy-Riemann equations
9m 3d du dv
— = — , and — = .
dx dv dy ox
3 See F. Morgan, Riemannian Geometry: A Beginner's Guide, 2nd ed. (A K Peters, Wellesley, 1998),
pp. 7-10. Figures 3.51 and 3.52 adapted with permission.
Miscellaneous Exercises for Chapter 3 243
(b) Show that u and v are harmonic, that is, that
(Also see §1.4 concerning the notion of torque.) Show
that
d2u d2U
9^ + 9^2
40. Suppose that a particle of mass m travels along a path
x according to Newton's second law F = ma, where
F is a gradient vector field. If the particle is also con-
strained to lie on an equipotential surface of F, show
that then it must have constant speed.
41 . Let a particle of mass m travel along a differentiable
path x in a Newtonian vector field F (i.e., one that
satisfies Newton's second law F = ma, where a is the
acceleration of x). We define the angular momen-
tum 1(f) of the particle to be the cross product of the
position vector and the linear momentum mv, that is,
l(r) = x(f) x m\(t).
(Here v denotes the velocity of x.) The torque about
the origin of the coordinate system due to the force F
is the cross product of position and force:
M(f ) = x(f) x F(f) = x(f) x ma(f).
dt
M.
Thus, we see that the rate of change of angular mo-
mentum is equal to the torque imparted to the particle
by the vector field F.
42. Consider the situation in Exercise 41 and suppose that
F is a central force (i.e., a force that always points
directly toward or away from the origin). Show that in
this case the angular momentum is conserved, that is,
that it must remain constant.
43. Can the vector field
F = (ex cos y + e~ x sin z) i — ex sin y j + e~x cos z k
be the gradient of a function f(x, y, z) of class C2?
Why or why not?
44. Can the vector field
F = *(y2 + \)i + (yex - ez) j + x2ez k
be the curl of another vector field G(x, y, z) of class
C2? Why or why not?
4 Maxima and Minima
in Several Variables
4.1 Differentials and Taylor's
Theorem
4.2 Extrema of Functions
4.3 Lagrange Multipliers
4.4 Some Applications of
Extrema
True/False Exercises for
Chapter 4
Miscellaneous Exercises
for Chapter 4
4.1 Differentials and Taylor's Theorem
Among all classes of functions of one or several variables, polynomials are without
a doubt the nicest in that they are continuous and differentiable everywhere and
display intricate and interesting behavior. Our goal in this section is to provide
a means of approximating any scalar-valued function by a polynomial of given
degree, known as the Taylor polynomial. Because of the relative ease with which
one can calculate with them, Taylor polynomials are useful for work in computer
graphics and computer-aided design, to name just two areas.
Taylor's Theorem in One Variable: A Review
Suppose you have a function /:lcR->R that is differentiable at a point a in
X. Then the equation for the tangent line gives the best linear approximation for
/ near a. That is, when we define p\ by
p\{x) = f(a) + f'(a)(x — a), we have pi(x) & fix) if x % a.
(See Figure 4.1.) As explained in §2.3, the phrase "best linear approximation"
means that if we take R\(x, a) to be f(x) — p\(x), then
lim
Ri(x, a)
0.
Note that, in particular, we have pi(a) = f(a) and p[(a) = f'(a).
Generally, tangent lines approximate graphs of functions only over very small
neighborhoods containing the point of tangency. For a better approximation, we
might try to fit a parabola that hugs the function's graph more closely as in
Figure 4.2. In this case, we want p2 to be the quadratic function such that
p2(a) = f(a), p'2(a) = f'(a), and p'2\a) = f"(a).
The only quadratic polynomial that satisfies these three conditions is
P2(x) = f(a) + f(a){x -a)+ ^(x - a)2.
It can be proved that, if / is of class C2, then
fix) = p2(x)+ R2(x, a),
4.1 | Differentials and Taylor's Theorem 245
where
Figure 4.3
fix) = In*
y = 2-
Approximations to
Riix, a)
lim = 0.
*-»-a ix — a)1
EXAMPLE 1 If fix) = In x, then, for a = 1 , we have
/(l) = lnl=0,
1
I
1
12
/'(I)
/"(I) =
1.
1.
Hence,
piix) = 0 + 1(*
p2(x) = 0+ 1(jc
1) = *- 1,
1) - I(* - l)2 =
2x
The approximating polynomials p\ and p2 are shown in Figure 4.3. ♦
There is no reason to stop with quadratic polynomials. Suppose we want to
approximate / by a polynomial of degree k, where k is a positive integer.
Analogous to the work above, we require that pk and its first k derivatives agree
with / and its first k derivatives at the point a. Thus, we demand that
Pk(a) =
f{a),
Pk(a) =
f'(a),
P'l(a) =
pf («) = /w(a).
Given these requirements, we have only one choice for pk, stated in the following
theorem:
- f(k)(
THEOREM 1.1 (Taylor's theorem in one variable) Let X be open in
R and suppose /:XcR^R is differentiable up to (at least) order k.
246 Chapter 4 I Maxima and Minima in Several Variables
Given a e X, let
f"(a) f(kXa)
pk(x) = f{a) + f'(a)(x -a)+ J-^-(x - af + • • • + J—^-{x - af. (1)
2 k\
Then
f(x) = Pk(x)+ Rk{x,a),
where the remainder term Rk is such that Rk(x, a)/(x — af -¥ 0 as x —* a.
Figure 4.4 The graphs of
(1) y =x -a,
(2) y = (x — a)2, and
(3) y = (x- af.
Note how much more closely the
graph of (3) hugs the jc-axis than
thatof(l)or(2).
The polynomial defined by formula (1) is called the &th-order Taylor poly-
nomial of / at a. The essence of Taylor's theorem is this: For x near a, the Taylor
polynomial pk approximates / in the sense that the error Rk involved in making
this approximation tends to zero even faster than (x — af does. When k is large,
this is very fast indeed, as we see graphically in Figure 4.4.
EXAMPLE 2 Consider In x with a = 1 again. We calculate
/(l) = lnl = 0,
/'(l) = y = l>
/"(I)
1
12
1*
i^l = (-l)k+\k - 1)!.
Therefore,
Pk(x) = (x
lf + -3(x
If
+
(-1)
<X ~ I f
Taylor's theorem as stated in Theorem 1.1 says nothing explicit about the
remainder term Rk. However, it is possible to establish the following derivative
form for the remainder:
PROPOSITION 1 .2 If / is of class Ck+\ then there exists some number z be-
tween a and x such that
Rk(x, a) = f( + )(Z\x - af+\
(Jfc + 1)!
(2)
In practice, formula (2) is quite useful for estimating the error involved with a
Taylor polynomial approximation. Both Theorem 1 . 1 (under the slightly stronger
hypothesis that / is of class Ck+ 1 ) and Proposition 1 .2 are proved in the addendum
to this section.
EXAMPLE 3 The fifth-order Taylor polynomial of f(x)= cos x about x = tt/2
is i « 1 <
/ 7T\ 1 / 7T\3 1 / 7T\5
P5(x) = -\x-2) + 6\x-2) -mV'V ■
4.1 | Differentials and Taylor's Theorem 247
(You should verify this calculation.) According to formula (2), the difference
between p5 and cos x is
R,
( n\ f(6\z) ( 11 \6 COSZ / 7t\(
\X'2) = ~^\X~2) =~~6T\X~2)
2/ 6! V 2/ 6! V 2.
where z is some number between n/2 and x. Since | cosjcI is never larger than 1,
we have
^ 7T\6 ^(jC-TT/2)6
*5 (*, |)
cos z
6!
Thus, for x in the interval [0, it], we have
(tt - 7T/2)6
R5 (x. |)
71'
720
46,080
720
0.0209.
In other words, the use of the polynomial ps above in place of cosx will be
accurate to at least 0.0209 throughout the interval [0, n]. ♦
Taylor's Theorem in Several Variables:
The First-order Formula
For the moment, suppose that /: X c R2 — > R is a function of two variables,
where X is open in R2 and of class Cl. Then near the point (a, b) e X, the best
linear approximation to / is provided by the equation giving the tangent plane at
(a, b, f(a, b)). That is,
f(x, y) » pi(x, v),
where
pi(x, y) = f(a, b) + fx(a, b)(x -a) + fy(a, b)(y - b).
Note that the linear polynomial p\ has the property that
pi(a, b) = f(a, b);
dpi df
-7— {a,b)= —(a,b),
ox ox
dpi df
-5— (a, b) = —(a, b).
oy oy
Such an approximation is shown in Figure 4.5.
To generalize this situation to the case of a function /: X c R" — > R of
class C1, we naturally use the equation for the tangent hyperplane. That is, if
z=f(x,y)
Figure 4.5 The graph of z = f(x, y) and
z = pi(x, y).
248 Chapter 4 I Maxima and Minima in Several Variables
a = (ai, a2, . . . , a„) e X, then
f(Xl,X2, ...,xn)& pi{xi,x2, ...,x„),
where
Pi(xi, xn) = /(a) + fxi(a)(xi - ai) + /X2(a)(x2 - a2)
+ r- fx„(a)(xn ~ an)-
Of course, the formula for p\ can be written more compactly using either
E -notation or, better still, matrices:
n
Pi(jci, . . . , xn) = /(a) + £ /*(»)(* - a,) = /(a) + D/(a)(x - a). (3)
; = 1
EXAMPLE 4 Let /(xj, X2, X3, X4) = xi + 2x2 + 3x3 + 4x4 + X1X2X3X4. Then
3/ 1 3/ „
= l + X2X3X4, = 2 + X1X3X4,
3xj 3x2
3/ „ 3/
- — = 3+xix2x4, - — =4 + X!X2x3.
dX3 dX4
At a = 0 = (0, 0, 0, 0), we have
^(0) = l, ^(0) = 2, |^(0) = 3, |^(0) = 4.
dx\ 6x2 0x3 dx4
Thus,
piixi, x2, x3, x4) = 0 + l(xi - 0) + 2(x2 - 0) + 3(x3 - 0) + 4(x4 - 0)
= X\ + 2X2 + 3X3 + 4X4.
Note that p\ contains precisely the linear terms of the original function /. On the
other hand, if a = (1 , 2, 3, 4), then
-Ml, 2, 3, 4) = 25, —^-(1,2, 3,4) = 14,
dxi dx2
df df
-Ml, 2, 3, 4)= 11, -Ml, 2, 3, 4) = 10,
dX3 dX4
so that, in this case,
pi(xi,X2,x3,x4) = 54 + 25(xi - 1)+ 14(x2 - 2) + ll(x3 -3)+ 10(x4 - 4).
♦
The relevant theorem regarding the first-order Taylor polynomial is just a re-
statement of the definition of differentiability. However, since we plan to consider
higher-order Taylor polynomials, we state the theorem explicitly.
THEOREM 1 .3 (First-order Taylor's formula in several variables) Let
X be open in R" and suppose that /:XcR"->R is differentiate at the point
a in X. Let
pi(x) = /(a) + D/(aXx-a)- (4)
Then
/(x) = pi(x) + *i(x, a),
where Ri(x, a)/||x — a|| — >• 0 as x —> a.
4.1 | Differentials and Taylor's Theorem 249
Note that we may also express the first-order Taylor polynomial using the gradient.
In place of (4), we would have
Pi(x) = /(a) + V/(a).(x-a).
Differentials
Before we explore higher-order versions of Taylor's theorem in several variables,
we consider the linear (or first-order) approximation in further detail.
Let h = x — a. Then formula (3) becomes
Pl(x) = /(a) + fl/(a)h = m + ■ (5)
i=i dx>
We focus on the sum appearing in formula (5) and summarize its salient
features as follows:
DEFINITION 1 .4 Let / : X C R" -> R and let ael The incremental
change of /, denoted Af, is
A/ = /(a + h)-/(a).
The total differential of /, denoted df(a, h), is
df df df
dX\ 0X2 dxn
The significance of the differential is that for h « 0,
Af « df.
(We have abbreviated df(a, h) by J /.)
Sometimes hj is replaced by the expression Ax, or dx{ to emphasize that it
represents a change in the z'th independent variable, in which case we write
At 9/ A A V A
df = —dxl + —dx2-\ h — dxn.
dx\ 0X2 6xn
(We've suppressed the evaluation of the partial derivatives at a, as is customary.)
EXAMPLE 5 Suppose f(x, y, z) = sin(xyz) + cos(xyz). Then
Af V j . &f , df
df = — dx + — dy + — dz
6x 6y 6z
= yz[cos(xyz) — sm(xyz)]dx + xz[cos(xyz) — sin(xyz)]dy
+ xy[cos(xvz) — sin(xyz)]dz
= (cos(xyz) — sin(xyz))(yz dx + xz dy + xy dz)- ^
The geometry of the differential arises, naturally enough, from tangent lines
and planes. (See Figures 4.6 and 4.7.) In particular, the incremental change Af
measures the change in the height of the graph of / when moving from a to a + h;
the differential change d f measures the corresponding change in the height of
250 Chapter 4 [ Maxima and Minima in Several Variables
Figure 4.6 The incremental
change A/ equals the change in
y -coordinate of the graph of
y = f(x) as the x-coordinate of a
point changes from a to a + dx.
The differential df equals the
change in y -coordinate of the
graph of the tangent line at a (i.e.,
the graph of y = p\(x)).
Figure 4.7 The incremental
change Af equals the change in
^-coordinate of the graph of
z = f(x, y) as a point in R2
changes from a = (a, b) to
a + h = (a + h, b + k). The
differential df equals the change in
z-coordinate of the graph of the
tangent plane at (a, b).
the graph of the (hyper)plane tangent to the graph at a. When h is small (i.e.,
when a + h is close to a), the differential df approximates the increment Af and
it is often easier from a technical standpoint to work with the differential.
EXAMPLE 6 Let /(*, y) = x - y + 2x2 + xy2. Then for (a, b) = (2,-1),
we have that the increment is
Af = f(2 + Ax, -1 + A.y) - f(2, -1)
= 2 + Ax - (-1 + Ay) + 2(2 + Ax)2 + (2 + Ax)(-1 + A.y)2 - 13
= lOAx - 5 Ay + 2(Ax)2 - 2Ax Ay + 2(A^)2 + Ax(Ay)2.
On the other hand,
df((2, -1), (Ax, Ay)) = fx(2, -I) Ax + fy(2, -I) Ay
= (1 + 4x + .v2)|(2,-i)Ax + (-1 + 2x>>)|(2,-i)A.y
= lOAx - 5A;y.
We see that df consists of exactly the terms of Af that are linear in Ax and Ay
(i.e., appear to first power only). This will always be the case, of course, since
that is the nature of the first-order Taylor approximation. Use of the differential
approximation is often sufficient in practice, for when Ax and Ay are small, higher
powers of them will be small enough to make virtually negligible contributions
to Af. For example, if Ax and Ay are both 0.01, then
df = (0.1 -0.05) = 0.05
and
Af = (0.1 - 0.05) + 0.0002 - 0.0002 + 0.0002 + 0.000001
= 0.05 + 0.000201 = 0.050201.
Thus, the values of df and Af are the same to three decimal places. ♦
4.1 | Differentials and Taylor's Theorem 251
EXAMPLE 7 A wooden rectangular block is to be manufactured with dimen-
sions 3 in x 4 in x 6 in. Suppose that the possible errors in measuring each di-
mension of the block are the same. We use differentials to estimate how accurately
we must measure the dimensions so that the resulting calculated error in volume
is no more than 0.1 in3 .
Let the dimensions of the block be denoted by x 3 in), y 4 in), and z
6 in). Then the volume of the block is
V=xyz and V « 3 ■ 4 • 6 = 72 in3.
The error in calculated volume is A V, which is approximated by the total differ-
ential dV. Thus,
A V w d V = Vx(3, 4, 6)Ax + Vy(3, 4, 6)Ay + V,(3, 4, 6)AZ
= 24Ax + 18Ay + 12Az.
If the error in measuring each dimension is e, then we have Ax = Ay = Az = e.
Therefore,
dV = 24Ax + 18Ay + 12Az = 24e + 18e + 12e = 54e.
To ensure (approximately) that | A V\ < 0.1, we demand
\dV\ = |54e| < 0.1.
Hence,
0.1
|e| < — = 0.0019 m.
" 54
So the measurements in each dimension must be accurate to within 0.0019 in. ♦
Figure 4.8 Which would
you buy?
EXAMPLE 8 The formula for the volume of a cylinder of radius r and height
h is V(r, h) = jtr2h. If the dimensions are changed by small amounts Ar and
Ah, then the resulting change A V in volume is approximated by the differential
change d V. That is,
AV ^dV =
dV dV ,
— Ar H Ah = InrhAr + nr Ah.
dr dh
Suppose the cylinder is actually a beer can, so that it has approximate dimensions
of r = 1 in and h = 5 in. Then
dV = jr(10Ar + Ah).
This statement shows that, for these particular values of r and h, the volume
is approximately 10 times more sensitive to changes in radius than changes in
height. That is, if the radius is changed by an amount e, then the height must be
changed by roughly lOe to keep the volume constant (i.e., to make AV zero).
We use the word "approximate" because our analysis arises from considering the
differential change d V rather than the actual incremental change A V.
This beer can example has real application to product marketing strategies.
Because the volume is so much more sensitive to changes in radius than height,
it is possible to make a can appear to be larger than standard by decreasing its
radius slightly (little enough so as to be hardly noticeable) and increasing the
height so no change in volume results. (See Figure 4.8.) This sensitivity analysis
shows that even a tiny decrease in radius can force an appreciable compensating
increase in height. The result can be quite striking, and these ideas apparently
252 Chapter 4 I Maxima and Minima in Several Variables
have been adopted by at least one brewery. Indeed this is how the author came to
fully appreciate differentials and sensitivity analysis.1 ♦
Taylor's Theorem in Several Variables:
The Second-order Formula
Suppose /: X c R2 —> R is a C2 function of two variables. Then we know that
the tangent plane gives rise to a linear approximation p\ of / near a given point
(a, b) of X. We can improve on this result by looking for the quadric surface that
best approximates the graph of z = fix, y) near (a, b, f(a, b)). See Figure 4.9
for an illustration. That is, we search for a degree 2 polynomial pi(x, y) = Ax2 +
Bxy + Cy2 + Dx + Ey + F such that, for (x, y) (a, b),
f(x, y) « p2(x, y).
Quadric
surface
Z=f(x,y)
Figure 4.9 The tangent plane and quadric
surface.
Analogous to the linear approximation pi, it is reasonable to require that p2 and
all of its first- and second-order partial derivatives agree with those of / at the
point (a, b). That is, we demand
dpi
dx
d2P2
dx2
(a, b) =
(a,b)
p2(a, b)
— (
dx
d2f
— — (
dx2
d2p2
(a,b),
(a, b),
2 (a, b)
f(a,b)
dp2
dy
d2P2
dxdy
d2f
(a, b) =
(a,b) =
df
dy
d2f
dxdy
{a, b),
(a,b),
(6)
dy
dy:
(a,b).
After some algebra, we see that the only second-degree polynomial meeting these
requirements is
P2(x, y) = f(a, b) + fx(a, b)(x -a) + fy(a, b)(y - b)
+ yxx(a, b)(x - a)2 + fxy(a, b)(x - a)(y - b)
+ \fyy(a,b)(y-b)2.
(7)
1 See S. J. Colley, The College Mathematics Journal, 25 (1994), no. 3, 226-227. Art reproduced with
permission from the Mathematical Association of America.
4.1 | Differentials and Taylor's Theorem 253
How does formula (7) generalize to functions of n variables? We need to begin
by demanding conditions analogous to those in (6) for a function /: X cR"->R.
For a = (a\, a2, . . . , an) £ X, these conditions are
p2(a) = /(a),
-^(a)=-^(a), i = l,2,...,n, (8)
dxi dxi
d2p2 d2f
(a)= — ^-(a), ij = 1,2,..., n.
dxidxj dxidxj
If you do some algebra (which we omit), you will find that the only polynomial
of degree 2 that satisfies the conditions in (8) is
n \ H
P2(x) = /(a) + A,(a)(*< " ad +2 E fxi*M(Xi ~ ai)(xJ - (9)
1=1 i.j=l
(Note that the second sum appearing in (9) is a double sum consisting of n2 terms.)
To check that everything is consistent when n = 2, we have
p2(xi,x2) = f{a\,a2) + fxM^' ai)(xi - fli) + fx2(ai,a2)(x2 - a2)
+ \ \_Sx^xSfl\^i){x\ - aif + fXlX2(ai, a2){x\ - a{)(x2 - a2)
+ fx2Xl(ai,a2)(x2 - a2){x\ - d\) + fXlX2{a\, a2)(x2 - a2)2] .
When /' is a C2 function, the two mixed partials are the same, so this formula
agrees with formula (7).
EXAMPLE 9 Let /(*, y, z) = ex+y+z and let a = (a, b, c) = (0, 0, 0). Then
/(0,0, 0) = e°= 1,
fx(0, 0, 0) = /,(0, 0, 0) = fz(0, 0, 0) = e° = 1,
fxx(Q, 0, 0) = fxy(0, 0, 0) = fxt(0, 0, 0) = fyy(0, 0, 0)
= fyz(0, 0, 0) = /„((), 0, 0) = e° = 1.
Thus,
P2(x, y,z) = l + l(x - 0) + l(y - 0) + l(z - 0)
+ \ [l(x - 0)2 + 2 ■ l(x - 0)(y - 0) + 2 ■ l(x - 0)(z - 0)
+ l(y - 0)2 + 2 • l(y - 0)(z - 0) + l(z - 0)2]
= 1 + x + y + z + k2 + xy + xz + + yz + \z2
= 1 + (x + y + z) + Ux + y + zf
2
We have made use of the fact that, since / is of class C , a term like
fxy(0, 0, 0)(x - 0)(y - 0) is equal to fyx(0, 0, 0)(y - 0)(x - 0).
Now we state the second-order version of Taylor's theorem precisely.
254 Chapter 4 I Maxima and Minima in Several Variables
THEOREM 1 .5 (Second-order Taylor's formula) Let X be open in R" , and
suppose that /:XcR"->R is of class C2. Let
n j n
P2OO = /(a) + fxi(&Xxi ~ ad + j Y fxixMXXi ~ a^XJ ~ ai)"
i=l Z <,J=1
Then
/(x) = p2(x) + fl2(x, a),
where |/?2|/||x - all2 -> 0 as x ->• a.
A version of Theorem 1.5, under the stronger assumption that / is of class C3, is
established in the addendum to this section.
EXAMPLE 10 Let f(x, y) = cosx cosy and (a, b) = (0, 0). Then
/(0,0) = 1;
fx(0, 0) = - sin x cosy I (0,0) = 0, fy(0, 0) = - cosx siny|(0,0) = 0;
fxx(0, 0) = - cosxcosy|(0 0) = -1,
fxy(Q, 0) = sinx siny|(0 0) = 0,
/yy(0,0) = - cosxcosy|(00) = -1.
Hence,
f(x, y) « p2(x, y) = 1 + \{-\ ■ x2 - 1 • y2) = 1 - \x2 - \y2 .
We can also solve this problem another way since / is a product of two functions.
We can multiply the two Taylor polynomials:
Pi(x, y) = (Taylor polynomial for cos x) • (Taylor polynomial for cos y)
= (1 - \x2) (1 - \y2)
= 1 — \x2 — \y2 up to terms of degree 2.
This method is justified by noting that if qi is the Taylor polynomial for cosine
and Ro is the corresponding remainder term, then
cosx cosy = [q2(x) + R2(x, 0)][q2(y) + R2(y, 0)]
= qi(x)q2(y) + q2(y)R2(x, 0) + q2(x)R2(y, 0) + R2(x, 0)R2(y, 0)
= q2(x)q2(y) + other stuff,
where (other stuff)/||(x, y)||2 —> 0 as (x, y) (0, 0), since both R2{x, 0) and
R2(y, 0) do. ♦
The Hessian
Recall that the formula for the first-order Taylor polynomial p\ was written quite
concisely in formula (5) by using vector and matrix notation. It turns out that it
is possible to do something similar for the second-order polynomial p2.
4.1 | Differentials and Taylor's Theorem 255
DEFINITION 1 .6 The Hessian of a function /:XcR"->R is the matrix
\x/,Vir\cf1 / t 1"Vi ptitr\7 ic f 1 r) v
YVllUov £ / 111 C11L1 y la u / / (JA
j OJij . 1 Hal la,
fx,X! fx,X2
fx\x„
Hf =
fx2X\ fx2x2
fx2x„
_ fx„xi fx„x2
fxnx„
The term "Hessian" comes from Ludwig Otto Hesse, the mathematician who
first introduced it, not from the German mercenaries who fought in the American
revolution.
Now let's look again at the formula for p2 in Theorem 1.5:
n j n
P2(x) = /(a) + £ fXi(*)hi + fx,* (»)/<;/',.
(We have let h = (hi, . . . , hn) = x — a.) This can be written as
p2« = /(a)+[/,1(a) /T2(a) ■■■ A„(a)]
&2
+
1 r
/i2 ■■• h„
Thus, we see that
/*2»(a) A2*2(a) ••• A2X„(a)
_/*„*, (a) /x„.v2(a) ••• A,,,, (a) J _
h2
p2(x) = /(a) + D/(a)h + ^hr///(a)h. (10)
(Remember that hT is the transpose of the w x 1 matrix h.)
EXAMPLE 11 (Example 10 revisited) For f(x, y) = cosx cos y, a = (0, 0),
we have
Df(x, y) = [— sinx cosy — cosx sin y]
and
Hf(x,y)
cos x cos y sin x sin y
sin x sin y — cos x cos y
256 Chapter 4 [ Maxima and Minima in Several Variables
Hence,
P2(x, y) = /(O, 0) + Df(0, 0)h + lhTHf(0, 0)h
l+[ 0 0]
+ \[h, h2]
' -1 0 "
" hi '
0 -1
= 1
l-h2
2 1
X-h2
2n2-
Once we recall that h = (hi, h2) = (x — 0, y — 0) = (x, y), we see that this result
checks with our work in Example 10, just as it should. ♦
Higher-order Taylor Polynomials
So far we have said nothing about Taylor polynomials of degree greater than 2
in the case of functions of several variables. The main reasons for this are (i) the
general formula is quite complicated and has no compact matrix reformulation
analogous to (10) and (ii) we will have little need for such formulas in this text.
Nonetheless, if your curiosity cannot be denied here is the third-order Taylor
polynomial for a function /: X c R" —> R of class C3 near aeX:
P3(x) = /(a) + fxi(*)(xi ~ ad + ~ ^ - atXxj - aj)
i=l i,j=\
1 "
i,j,k=l
(The relevant theorem regarding pi is that f(x) — pj,(x) + Rj,(x, a), where
|/?3(x, a)|/||x — a || 3 -> 0 as x -> a.) If you must know even more, the /cth-order
Taylor polynomial is
i=l Z i,y=l
1 "
H h jti XI A (a^Xi'i ~ fl,'i) ' ' ' ^ ~ a<* )•
l'l,...,H = l
Formulas for Remainder Terms (optional)
Under slightly stricter hypotheses than those appearing in Theorems 1.3 and
1.5, integral formulas for the remainder terms may be derived as follows. Set
h = x — a. If / is of class C2, then
n p 1
Ri(x, a) = V / (1 - / )/, ,, (a + th)h,hj dt
= [ [hT#/(a + fh)h](l -t)dt.
Jo
4.1 | Differentials and Taylor's Theorem 257
If / is of class C3, then
v r<i2
^ (l - o2
R2(x, a) = 2_, / — ^ — ./',,«,, (a + th)hihjhk dt,
and if / is of class C*+1, then
i?,(x,a)= J] / — — — fxiyxi2-xit+1(a + t\L)hhhi2 ■ ■ ■ hh+l dt.
U '*+i=i Jo
Although explicit, these formulas are not very useful in practice. By artful appli-
cation of Taylor's formula for a single variable, we can arrive at derivative versions
of these remainder terms (known as Lagrange's form of the remainder) that are
similar to those in the one-variable case.
Lagrange's form of the remainder. If / is of class C2, then in Theorem 1 .3
the remainder Ri is
1 "
#i(x, a) = - J] fXiXj(z)hihj
2 UM
for a suitable point z in the domain of / on the line segment joining a and
x = a + h. Similarly, if / is of class C3, then the remainder R2 in Theorem
1.5 is
1 -
tf2(x, a)= — fXiXjXk{z)hihjhk
31
ij,k=l
for a suitable point z on the line segment joining a and x = a + h. More
generally, if / is of class Ck+l, then the remainder Rk is
*^X' a) = TkTiv fw-*** V)KK ■ ■ ■ K+i
K <i,-,i*+i=1
for a suitable point z on the line segment joining a and x = a + h.
The remainder formulas above are established in the addendum to this section.
EXAMPLE 12 For fix, y) = cos x cos y, we have
2
X] fxiXjXk(z)hihjhk
\R2(x,y,0, 0)| = i
2
<
tj,*=i
since all partial derivatives of / will be a product of sines and cosines and, hence,
no larger than 1 in magnitude. Expanding the sum, we get
\R2(x,y,0, 0)| < i(|/Jl|3 + 3/i?|/!2| + 3|/i1|^ + |/z2|3).
If both \hi \ and \ fi2\ are no more than, say, 0.1, then
\R2(x,y,0, 0)| < i (8- (O.l)3) = 0.0013.
Chapter 4 i Maxima and Minima in Several Variables
y
z
0.1
Z = cos x cos y
-0.1
0.1
X
y
-0.1
X
Figure 4.11 The graph of f(x, y) =
cos x cos y and its Taylor polynomial
Figure 4.10 The polynomial p2
approximates / to within 0.0013
on the square shown. (See
Example 12.)
Pi(x, y) = 1 — \x2 — \y2 over the square
{(Jt.y) 1-1 <*<!,-! <?<!}.
So throughout the square of side 0.2 centered at the origin and shown in Fig-
ure 4.10, the second-order Taylor polynomial is accurate to at least 0.0013 (i.e., to
two decimal places) as an approximation of f(x, y) = cosx cosy. In Figure 4. 11,
we show the graph of f(x, y) — cos x cos y over the square domain {(x , y) | — 1 <
x < 1 , — 1 < y < 1} together with the graph of its second-order Taylor polyno-
mial p2(x, y) — 1 — \x2 — \y2 (calculated in Example 10). Note how closely the
surfaces coincide near the point (0, 0, 1), just as the analysis above indicates. ♦
Addendum: Proofs of Theorem 1.1, Proposition 1.2,
and Theorem 1.5
Below we establish some of the fundamental results used in this section. We begin
by proving Theorem 1.1, Taylor's theorem for function of a single variable, and
Proposition 1 .2 regarding the remainder term in Theorem 1.1. We then use these
results to "bootstrap" a proof of the multivariable result of Theorem 1.5 and to
derive Lagrange's formula for the remainder term appearing in it.
Proof of Theorem 1 .1 We prove the result under the stronger assumption that /
is of class Ck+l rather than assuming that / is only differentiate up to order k.
(This distinction matters little in practice.)
By the fundamental theorem of calculus,
We evaluate the integral on the right side of ( 1 1 ) by means of integration by parts.
Recall that the relevant formula is
We use this formula with u = f'(t) and v = x — t so that dv = —dt. (Note that
in the right side of (1 1), x plays the role of a constant.) We obtain
(11)
u dv = uv —
(12)
4.1 | Differentials and Taylor's Theorem 259
Combining (11) and (12), we have
fix) = f(a) + f'(a)(x -a) + \\x - t)f"(t)dt. (13)
Thus, we have shown, when / is differentiable up to (at least) second order, that
Ri(x,a)= f (x - t)f"(t)dt.
J a
This provides an integral formula for the remainder in formula ( 1 ) of Theorem 1 . 1
when k = 1, although we have not yet established that R\(x,a)/(x — a) — > 0 as
x — > a.
To obtain the second-order formula, the case k = 2 of (1), we focus on
Ri(x, a) = f*(x — t)f"{t)dt and integrate by parts again, this time with u =
f"(t) and v = (x - tf/2, so that dv = -(x - t)dt. We obtain
f\x
J a
t)f"(t)dt =
f"(t)(x - tf
f"(a)(x - a)2
a Ja
f
f"'(t)dt
(x - tf
f"'(t)dt.
Hence (13) becomes
f"(a)
f(x) = f(a) + f'(a)(x -a)+ J—^-{x
2 r* (x-t)
a) +L —
Therefore, we have shown, when / is differentiable up to (at least) third order,
that
-Jlf"\t)dt.
R2(x,a)= f
J a
We can continue to argue in this manner or use mathematical induction to show
that formula (1) holds in general with
Rk(x,a) = j"^t^\t)dt, (14)
assuming that / is differentiable up to order (at least) k+\.
It remains to see that Rk(x, a)/(x — a)k — > 0 as x — > a. In formula (14) we
are only considering t between a and x, so that \x — t \ < \x — a\. Moreover, since
we are assuming that / is of class Ck+1, we have that f(k+x\t) is continuous and
therefore, bounded for t between a and x (i.e., that \f{k+l\t)\ < M for some
constant M). Thus,
\Rk(x,a)\ <
r (x - tf
Ja k\
f(k+\t)dt
< ±
f
(x - tf
k\
fik+l)(t)
dt,
where the plus sign applies if x > a and the negative sign if x < a,
Cx M
x - a\k dt
M
\k+\
Thus,
Rk(x, a)
(x — a)k
M
as x —> a, as desired.
260 Chapter 4 I Maxima and Minima in Several Variables
Proof of Proposition 1.2 We establish Proposition 1.2 by means of a general
version of the mean value theorem for integrals. This theorem states that for
continuous functions g and h such that h does not change sign on [a, b] (i.e.,
either h(t) > 0 on [a, b] or h(t) < 0 on [a, b]), there is some number z between
a and b such that
f g(t)h(t)dt = g(z) f h(t)dt.
(We omit the proof but remark that this theorem is a consequence of the interme-
diate value theorem.) Applying this result to formula (14) with g(t) = f(k+^(t)
and h(t) = (x — t)k /k\, we find that there must exist some z between a and x
such that
i) = f(k+\z)\X{^-^dt=f^\z)
k\
(x - t)
(k+iy.
(k + iy.
(x — a)
k+l
Proof of Theorem 1.5 As in the proof of Theorem 1 . 1 , we establish Theorem 1 .5
under the stronger assumption that / is of class C3 . Begin by setting h = x — a,
so that x = a + h, and consider a and h to be fixed. We define the one-variable
function F by F(t) = /(a + th). Since / is assumed to be of class C3 on an
open set X, if we take x sufficiently close to a, then F is of class C3 on an open
interval containing [0, 1]. Thus, Theorem 1.1 with k = 2, a = 0, and x = 1 may
be applied to give
F(l) = F(0) + F'(0)(1 - 0) + ^9(1 - 0)2 + R2(l, 0)
F"(0)
= F(0) + F'(0) + — ^ + R2(l, 0),
(15)
where R2(l, 0) = /J ^-F"'(0^.Now we use the chain rule to calculate deriva-
tives of F in terms of partial derivatives of /:
.7 = 1
hi = ^ fXiXj(a + th)hihj;
F'(t) = D/(a + th)h = J2 U ■(» + fh)^
F"(r) = £
F"'(t) =
k=\
Thus, (15) becomes
n j n
/(a + h) = /(a) + fx,Wi + ~ E UM)hihJ
J2 /u,,,;(a • /h)/;,/;,
.W=i
fyfc = X! fxiXjxM + t^hihjhk
i,j,k=\
+
n 1
(l - o2
fXiXjXk(a + th)hihjhkdt,
4.1 | Differentials and Taylor's Theorem 261
or, equivalently,
n \ n
/(x) = /(a) + /*,(»)(*; " ad +~Y1 fxitjW&i ~ ai)(xi ~ ai)
+ #2(x, a),
where the multivariable remainder is
R2(x,a)= V / y——LfXiX]Xl(a + th)hihjhkdt.
>,j,k=\ Jo l
(16)
We must still show that |i?2(x, a)|/||x — a||2 -> 0 as x -> a, or, equivalently,
that |^?2(x, a)|/|h||2 -> 0 as h -> 0. To demonstrate this, note that, for a and h
fixed the expression (1 — t)2 fXjXjXk(st + fh) is continuous for f in [0, 1] (since /
is assumed to be of class C3), hence bounded. In addition, for i = 1, . . . , n, we
have that |/z,-| < ||h||.
Hence,
|tf2(x,a)| =
<
;i 1
<
n «1
(i - o2
2
(1 " tf
fxtXjxM + th)hihjhk dt
fXiXjXt(a + th)hjhjhk dt
M||/i||3 dt = n3M||h||3 = n3M||x - a|
Thus,
|i?2(x, a)| ,
v '1 <«3M||x-a|
x- a
0
as x -> a.
Finally, we remark that entirely similar arguments may be given to establish
results for Taylor polynomials of orders higher than two. ■
Lagrange's formula for the remainder (see page 257) Using the function
F(t) = /(a + fh) defined in the proof of Theorem 1.5, Proposition 1.2 implies
that there must be some number c between 0 and 1 such that the one-variable
remainder is
F"'(c) -
tf2(l,0)=^-Al-0)3.
Now, the remainder term Rz(I, 0) from Proposition 1.2 is precisely ^(x, a) in
Theorem 1.5 and
n n
F"'(c)= fXiXiXt{* + c\v)hihjhk= fXiXjXk(z)hihjhk,
i,j,k=\ i,j,k=l
where z = a + ch. Since c is between 0 and 1 , the point z lies on the line segment
joining a and x = a + h, and so
1
^2(x, a)=— fXiXjXk(z)hihjhk
i,j,k=l
which is the result we desire. The derivation of the formula for Rk(x, a) for k > 2
is analogous. ■
262 Chapter 4 I Maxima and Minima in Several Variables
4.1 Exercises
In Exercises 1—7, find the Taylor polynomials of given order
k at the indicated point a.
1. f(x) = e2x,a = 0,k = A
2. fix) = ln(l + x), a = 0,k = 3
3. f(x)= l/x2,a = l,k = 4
4. f(x) = Jx,a = l,k = 3
5. f(x) = </x, a = 9, k = 3
6. f(x) = sinx, a = 0, k = 5
7. f(x) = sinx, a = n/2, k = 5
In Exercises 8— 15, find the first- and second-order Taylor poly-
nomials for the given function f at the given poin t a.
8. /(x,y) = l/(x2 + y2 + l),a = (0,0)
9. f(x, y) = l/(x2 + y2 + 1), a = (1, -1)
10. f(x,y) = e2x+y,a = (0, 0)
11. f(x, y) = e2x cos3y, a = (0, jr)
12. /(x, y, z) = ye3x + ze2-\ a = (0, 0, 2)
13. /(jc, y, z) = xy- 3y2 + 2xz, a = (2, -1, 1)
14. /(x, z) = l/(x2 + y2 + z2 + 1), a = (0, 0, 0)
15. f(x, y, z) = sinxyz, a = (0, 0, 0)
In Exercises 16-20, calculate the Hessian matrix ///(a) for
the indicated function f at the indicated point a.
16. f(x,y)= l/(x2 + y2+l),a = (0,0)
17. /(x, y) = cosx siny, a = (tt/4, it/3)
18. f(x,y,z)= -£=,a = (l,2,-4)
19. /(x, y, z) = x3 + x2y - yz2 + 2z\ a = (1, 0, 1)
20. f(x, y, z) = e2x~^ sin5z, a = (0, 0, 0)
21 . For / and a as given in Exercise 8, express the second-
order Taylor polynomial pi(x, y), using the derivative
matrix and the Hessian matrix as in formula (10) of
this section.
22. For / and a as given in Exercise 1 1 , express the second-
order Taylor polynomial pi{x, y), using the derivative
matrix and the Hessian matrix as in formula (10) of
this section.
23. For / and a as given in Exercise 12, express the second-
order Taylor polynomial p2(x, y, z), using the deriva-
tive matrix and the Hessian matrix as in formula (10)
of this section.
24. For / and a as given in Exercise 1 9, express the second-
order Taylor polynomial pi(x, y, z), using the deriva-
tive matrix and the Hessian matrix as in formula (10)
of this section.
25. Consider the function
f(xux2,...,x,,) = ex>+2x>+-+"x».
(a) Calculate D/(0, 0, . . . , 0) and Hf(0, 0, . . . , 0).
(b) Determine the first- and second-order Taylor poly-
nomials of / at 0.
(c) Use formulas (3) and ( 10) to write the Taylor poly-
nomials in terms of the derivative and Hessian
matrices.
26. Find the third-order Taylor polynomial pi(x, y, z) of
f(x,y,z) = ex+2y+3z
at (0, 0, 0).
27. Find the third-order Taylor polynomial of
f{x, y, z) = x4 + x3y + 2y3 - xz2 + x2y + 3xy - z + 2
(a) at (0, 0, 0).
(b) at (1,-1,0).
Determine the total differential of the functions given in
Exercises 28—32.
28. /(x,y) = x2y3
29. f{x, y, z) = x2 + 3y2 - 2z3
30. fix, y, z) = cos (xyz)
31. fix, y, z) = ex cosy + ey sin z
32. fix, y, z) = l/^Jxyz
33. Use the fact that the total differential df approximates
the incremental change A/ to provide estimates of the
following quantities:
(a) (7.07)2(1.98)3
(b) l/y(4.1)(1.96)(2.05)
(c) (1. 1) cos ((jr -0.03X0.12))
34. Near the point (1, —2, 1), is the function g(x, y, z) =
x3 — 2xy + x2z + 7z most sensitive to changes in x,
y, or z?
35. To which entry in the matrix is the value of the
determinant
2 3
-1 5
most sensitive?
4.2 | Extrema of Functions 263
36. If you measure the radius of a cylinder to be 2 in, with
a possible error of ±0. 1 in, and the height to be 3 in,
with a possible error of ±0.05 in, use differentials to
determine the approximate error in
(a) the calculated volume of the cylinder.
(b) the calculated surface area.
37. A can of mushrooms is currently manufactured to have
a diameter of 5 cm and a height of 12 cm. The man-
ufacturer plans to reduce the diameter by 0.5 cm. Use
differentials to estimate how much the height of the
can would need to be increased in order to keep the
volume of the can the same.
38. Consider a triangle with sides of lengths a and b that
make an interior angle 0 .
(a) If a = 3, b = 4, and 0 = tt/3, to changes in which
of these measurements is the area of the triangle
most sensitive?
(b) If the length measurements in part (a) are in error
by as much as 5% and the angle measurement is
in error by as much as 2%, estimate the resulting
maximum percentage error in calculated area.
39. To estimate the volume of a cone of radius approx-
imately 2 m and height approximately 6 m, how ac-
curately should the radius and height be measured so
that the error in the calculated volume estimate does
not exceed 0.2 m3? Assume that the possible errors in
measuring the radius and height are the same.
40. Suppose that you measure the dimensions of a block
of tofu to be (approximately) 3 in by 4 in by 2 in.
Assuming that the possible errors in each of your mea-
surements are the same, about how accurate must your
measurements be so that the error in the calculated
volume of the tofu is not more than 0.2 in3? What per-
centage error in volume does this represent?
41 . (a) Calculate the second-order Taylor polynomial for
f(x, y) = cosx siny at the point (0, ?r/2).
(b) If h = (huh2) = (x, y) - (0, tt/2) is such that
\hi \ and \h2\ are no more than 0.3, estimate how
accurate your Taylor approximation is.
42. (a) Determine the second-order Taylor polynomial of
f(x, y) = ex+2y at the origin.
(b) Estimate the accuracy of the approximation if \x\
and \ y\ are no more than 0.1.
43. (a) Determine the second-order Taylor polynomial of
f(x, y) = e2x cos y at the point (0, rr/2).
(b) If h = (huh2) = (x, y) - (0, tt/2) is such that
\h\\ < 0.2 and \h2\ < 0.1, estimate the accuracy
of the approximation to / given by your Taylor
polynomial in part (a).
4.2 Extrema of Functions
The power of calculus resides at least in part in its role in helping to solve a wide
variety of optimization problems. With any quantity that changes, it is natural to
ask when, if ever, does that quantity reach its largest, its smallest, its fastest or
slowest? You have already learned how to find maxima and minima of a function
of a single variable, and no doubt you have applied your techniques to a number of
situations. However, many phenomena are not appropriately modeled by functions
of only one variable. Thus, there is a genuine need to adapt and extend optimization
methods to the case of functions of more than one variable. We develop the
necessary theory in this section and the next and explore a few applications in §4.4.
Critical Points of Functions
Let X be open in R" and /:XcR"^Ra scalar- valued function.
DEFINITION 2.1 We say that / has a local minimum at the point a in
X if there is some neighborhood U of a such that f(x) > /(a) for all x
in U. Similarly, we say that / has a local maximum at a if there is some
neighborhood U of a such that /(x) < /(a) for all x in U.
Figure 4.1 2 The graph of When n = 2, local extrema of f(x, y) are precisely the pits and peaks of the
z = f(x, y). surface given by the graph of z = f(x, y), as suggested by Figure 4.12.
Max.
264 Chapter 4 I Maxima and Minima in Several Variables
We emphasize our use of the adjective "local." When a local maximum of
a function / occurs at a point a, this means that the values of / at points near
a can be no larger, not that all values of / are no larger. Indeed, / may have
local maxima and no global (or absolute) maximum. Consider the graphs in
Figure 4.13. (Of course, analogous comments apply to local and global minima.)
Figure 4.1 3 Examples of local and global maxima.
Recall that, if a differentiable function of one variable has a local extremum
at a point, then the derivative vanishes there (i.e., the tangent line to the graph
of the function is horizontal). Figures 4.12 and 4.13 suggest strongly that, if a
function of two variables has a local maximum or minimum at a point in the
domain, then the tangent plane at the corresponding point of the graph must be
horizontal. Such is indeed the case, as the following general result (plus formula
(4) of §2.3) implies.
THEOREM 2.2 Let X be open in R" and let /:XcR"^R be differentiable.
If / has a local extremum at a 6 X, then Df(a) = 0.
PROOF Suppose, for argument's sake, that / has a local maximum at a. Then the
one-variable function F defined by F(t) = /(a + 1 h) must have a local maximum
at t = 0 for any h. (Geometrically, the function F is just the restriction of / to the
line through a parallel to h as shown in Figure 4. 14.) From one-variable calculus,
we must therefore have F'(0) = 0. By the chain rule
F'(t) = -[/(a + th)] = D/(a + th)h = V/(a + th) • h.
at
1 \
1 k
1 / \
1 1
1 / 1
1
1 J
Graph of /
restricted to line
Figure 4.1 4 The graph of / restricted to a line.
4.2 | Extrema of Functions 265
Hence,
/-o
X/<o
/ \
/ \
1 \
f-o
v/
/\
/ \
/f>0\
\ f>0/
\ 1
\ /
\ /
\ /
\ /
\ /
Figure 4.1 5 The function / is
strictly positive on the shaded
region, strictly negative on the
unshaded region, and zero along
the lines y = ±x.
0 = F'(0) = D/(a)h = fx^)hx + /a(a)A2 + • • • + A„(a)fc„.
Since this last result must hold for all h e R", we find that by setting h in turn
equal to (1, 0, . . . , 0), (0, 1, 0, . . . , 0), . . . , (0, . . . , 0, 1), we have
fXl{*)= = -•• = /*„ (a) = 0.
Therefore, Df(a) = 0, as desired. ■
A point a in the domain of / where Df(a) is either zero or undefined is called
a critical point of /. Theorem 2.2 says that any extremum of / must occur at a
critical point. However, it is by no means the case that every critical point must
be the site of an extremum.
EXAMPLE 1 If f(x, y) = x2 - y2, then Df(x, y) = [ 2x -2y ] so that,
clearly, (0, 0) is the only critical point. However, neither a maximum nor a mini-
mum occurs at (0, 0). Indeed, inside every open disk centered at (0, 0), no matter
how small, there are points for which f(x, y) > f(0, 0) = 0 and also points where
f(x, y) < /(O, 0). (See Figure 4.15.) ♦
This type of critical point is called a saddle point. Its name derives from the
fact that the graph of z = f(x , y) looks somewhat like a saddle. (See Figure 4.16.)
Figure 4.1 6 A saddle point.
EXAMPLE 2 Let f(x, y) = ^x1 + y2. The domain off is all of R2. We com-
2x 2y
pute that Df(x, y)
; note that Df is unde-
3(x2 + y2)2/3 3(x2 + _y2)2/3 _
fined at (0, 0) and nonzero at all other (x, y) e R2. Hence, (0, 0) is the only
critical point. Since f(x, y) > 0 for all (x, y) and has value 0 only at (0, 0), we
see that / has a unique (global) minimum at (0, 0). ♦
The Nature of a Critical Point: The Hessian Criterion
We illustrate our current understanding regarding extrema with the following
example:
EXAMPLE 3 We find the extrema of
f(x, y) = x2 + xy + y2 + 2x- 2y + 5.
266 Chapter 4 | Maxima and Minima in Several Variables
/(«)
Figure 4.17 An
upward-opening parabola.
Figure 4.18 A
downward-opening parabola.
Since / is a polynomial, it is differentiable everywhere, and Theorem 2.2 implies
that any extremum must occur where df/dx and df/dy vanish simultaneously.
Thus, we solve
Bf
dx
Bf
dy
= 2x + y + 2 = 0
= x+2y-2 = 0
and find that the only solution is x = —2, y = 2. Consequently, (—2, 2) is the
only critical point of this function.
To determine whether (—2, 2) is a maximum or minimum (or neither), we
could try graphing the function and drawing what we hope would be an obvious
conclusion. Of course, such a technique does not extend to functions of more than
two variables, so a graphical method is of limited value at best. Instead we'll see
how / changes as we move away from the critical point:
A/ = /(-2 + A,2 + *)-/(-2,2)
= [(-2 + hf + (-2 + h)(2 + k) + {2 + kf
+ 2(-2 + /!)-2(2 + £) + 5]- 1
= h2 + hk + k2.
If the quantity Af = h2 + hk + k2 is nonnegative for all small values of h and k,
then (—2, 2) yields a local minimum. Similarly, if Af is always nonpositive, then
(—2, 2) must yield a local maximum. Finally, if Af is positive for some values
of h and k and negative for others, then (—2, 2) is a saddle point. To determine
which possibility holds, we complete the square:
Af
h2 + hk + k2
h2 + hk +
,k + . k
(h + \kf + Ik2.
Thus, Af > 0 for all values of h and k, so (—2, 2) necessarily yields a local
minimum. ♦
Example 3 with its attendant algebra clearly demonstrates the need for a better
way of determining when a critical point yields a local maximum or minimum (or
neither). In the case of a twice differentiable function /:XcR^R, you already
know a quick method namely, consideration of the sign of the second derivative.
This method derives from looking at the second-order Taylor polynomial of /
near the critical point a, namely,
f"(a) ,
f(x) « p2(x) = f(a) + / (a)(x -a)-\ — (x - a)
= /(a)H ^—(x ~ a) ,
since /' is zero at the critical point a of /. If f"{a) > 0, the graph of y = pi(x)
is an upward-opening parabola, as in Figure 4.17, whereas if f"{a) < 0, then
the graph of y = pi(x) looks like the one shown in Figure 4.18. If f"(a) = 0,
then the graph of y = pi(x) is just a horizontal line, and we would need to use
a higher-order Taylor polynomial to determine if / has an extremum at a. (You
may recall that when f"(a) = 0, the second derivative test from single-variable
calculus gives no information about the nature of the critical point a.)
The concept is similar in the context of n variables. Suppose that
/(*)= f(xi,x2, . . . ,xn)
4.2 | Extrema of Functions 267
is of class C2 and that a = (a\, a2, . . . , a„) is a critical point of /. Then the
second-order Taylor approximation to / gives
a/ = /(x) - m « P2(x) - m
= Z)/(a)(x - a) + i(x - a)r///(a)(x - a)
when x a. (See Theorem 1.5 and formula (10) in §4.1.) Since / is of class C2
and a is a critical point, all the partial derivatives vanish at a, so that we have
D/(a) = 0 and hence,
A/«i(x-a)r///(a)(x-a).
(1)
The approximation in (1) suggests that we may be able to see whether the in-
crement A / remains positive (respectively, remains negative) for x near a and
hence, whether / has a local minimum (respectively, a local maximum) at a by
seeing what happens to the right side.
Note that the right side of (1), when expanded, is quadratic in the terms
(xj — cii). More generally, a quadratic form in h\, h2, ■ . . , h„ is a function Q
that can be written as
n
Q(hu h2, hn) = ^ bijhihj,
where the bij 's are constants. The quadratic form Q can also be written in terms
of matrices as
Q(h)=[hl h2
~ bu
b\2 ■
■ bl„
~ hx ~
h„]
b2\
b22 ■
■ b2n
h2
_ K\
bnl ■
■ h
unn -
_ K _
hrfih,
(2)
where B = (pij). Note that the function Q is unchanged if we replace all bjj with
^{bij + bjj). Hence, we may always assume that the matrix B associated to Q is
symmetric, that is, that bjj = bj, (or, equivalently, that BT = B). Ignoring the
factor of 1 /2, we see that the right side of (1) is the quadratic form in h = x — a,
corresponding to the matrix B = Hf(a).
A quadratic form Q (respectively, its associated symmetric matrix B) is said
to be positive definite if Q(h) > 0 for all h ^ 0 and negative definite if Q(h) < 0
for all h ^ 0. Note that if Q is positive definite, then Q has a global minimum (of
0) at h = 0. Similarly, if Q is negative definite, then Q has a global maximum at
h = 0.
The importance of quadratic forms to us is that we can judge whether / has
a local extremum at a critical point a by seeing if the quadratic form in the right
side of ( 1) has a maximum or minimum at x = a. The precise result, whose proof
is given in the addendum to this section, is the following:
THEOREM 2.3 Let U c R" be open and /: U R a function of class C2.
Suppose that a £ U is a critical point of /.
1. If the Hessian Hf(a) is positive definite, then / has a local minimum at a.
2. If the Hessian Hf(a) is negative definite, then / has a local maximum at a.
3. If det Hf(a) 0 but Hf(a) is neither positive nor negative definite, then /
has a saddle point at a.
268 Chapter 4 I Maxima and Minima in Several Variables
In view of Theorem 2.3, the issue thus becomes to determine when the Hessian
Hf(a) is positive or negative definite. Fortunately, linear algebra provides an
effective means for making such a determination, which we state without proof.
Given a symmetric matrix B (which, as we have seen, corresponds to a quadratic
form Q), let B^, for k = 1, . . . , n, denote the upper leftmost k x k submatrix of
B. Calculate the following sequence of determinants:
bn b22
det 5,
det B7
det B3 =
bn bi2 bn
bi\ bi2 bn
b-i\ bi2 b3i
det£„ = det 5.
If this sequence consists entirely of positive numbers, then B and Q are positive
definite. If this sequence is such that det B^ < 0 for k odd and det B^ > 0 for k
even, then B and Q are negative definite. Finally, if det B ^ 0, but the sequence
of determinants det B\ , det B2, . . . , det B„ is neither of the first two types, then B
and Q are neither positive nor negative definite. Combining these remarks with
Theorem 2.3, we can establish the following test for local extrema:
Second derivative test for local extrema. Given a critical point a of a func-
tion / of class C2, look at the Hessian matrix evaluated at a:
tf/(a)
/*,*,(») fxlX2(*) ■■■ /*,*„(»)
fX2xM) fx2x2(a) ••• /x2x„(a)
From the Hessian, calculate the sequence of principal minors of Hf(a).
This is the sequence of the determinants of the upper leftmost square sub-
matrices of Hf(a). More explicitly, this is the sequence d\, d2, . . . , dn,
where dt = det Ht, and Hk is the upper leftmost k x k submatrix of Hf(a).
That is,
d\
di
/x,x,(a),
/««(»)
/v!*2(a)
fx2xM)
fx2x2(&)
1
fxixM)
fxlxM
fxlxM
fx2xM)
fx2xM
fx2xM
/v3*,(a)
/v3*2(a)
inu(a)
di= fx2xM) fx2x2(a) fx2x3(a) , ...,d„ = \Hf(a)\.
The numerical test is as follows:
Assume that d„ = det Hf(a) ^ 0.
1. If dk > 0 for k = 1 , 2, . . . , n, then / has a local minimum at a.
2. If dk < 0 for k odd and dk > 0 for k even, then / has a local maximum
at a.
3. If neither case 1 nor case 2 holds, then / has a saddle point at a.
In the event that det ///(a) = 0, we say that the critical point a is degenerate
and must use another method to determine whether or not it is the site of an
extremum of /.
4.2 | Extrema of Functions 269
EXAMPLE 4 Consider the function
fxx
fxy
2 1
. fyx
fyy _
1 2
minors
is d
i = /«(-2, 2)
/(*, y) = x2 + xy + y2 + 2x - 2y + 5
in Example 3. We have already seen that (—2, 2) is the only critical point. The
Hessian is
Hf(x, y) =
\Hf(—2, 2)| = 3 (> 0). Hence, / has a minimum at (—2, 2), as we saw before,
but this method uses less algebra. ♦
EXAMPLE 5 (Second derivative test for functions of two variables) Let us
generalize Example 4. Suppose that f(x, y) is a function of two variables of class
C2 and further suppose that / has a critical point at a = (a, b). The Hessian
matrix of / evaluated at {a, b) is
Hf(a,b)
fxx(a,b) fxy(a,b)
fxy(a,b) fyy(a,b)
Note that we have used the fact that fxy = fyx (since / is of class C2) in con-
structing the Hessian. The sequence of principal minors thus consists of two
numbers:
d\ = fxx(a,b) and d2 = fxx(a, b)fyy(a, b) - fxy(a, bf .
Hence, in this case, the second derivative test tells us that
1. / has a local minimum at (a, b) if
fxx(a, b) > 0 and fxx(a, b)fyy(a, b) - fxy(a, bf > 0.
2. / has a local maximum at {a, b) if
fxx(a, b) < 0 and fxx(a, b)fyy(a, b) - fxy(a, bf > 0.
3. / has a saddle point at (a, b) if
fxxia, b)fyy(a, b) - fxy(a, bf < 0.
Note that if fxx(a, b)fyy(a, b) — fxy(a, bf = 0, then / has a degenerate critical
point at (a, b) and we cannot immediately determine if {a, b) is the site of a local
extremum of /. ♦
EXAMPLE 6 Let f(x, y, z) = x3 + xy2 + x2 + y2 + 3z2. To find any local
extrema of /, we must first identify the critical points. Thus, we solve
Df(x,y,z)= [3x2 + y2 + 2x 2xy + 2y 6z] = [0 0 0] .
From this, it is not hard to see that there are two critical points: (0, 0, 0) and
(-§, 0,0). The Hessian of/ is
Hf(x,y,z) =
6x + 2
2y
o
2y
2x + 2
0
270 Chapter 4 I Maxima and Minima in Several Variables
At the critical point (0, 0, 0), we have
Hf(0, 0, 0) =
2 0 0
0 2 0
0 0 6
and its sequence of principal minors is d\ =2, d2 = 4, dj = 24. Since these
determinants are all positive, we conclude that / has a local minimum at (0,0,0).
At (-§,0,0), we calculate that
ff/(--,0,0
-2 0 0
0 | 0
0 0 6
The sequence of minors is —2, — |, — 8. Hence, / has a saddle point at (— |, 0, 0).
+
h>0
+
+
+
1 —
-1
0
+
1
1
+
+
+
Figure 4.1 9 Away from the
origin, the function h of Example 7
is negative along the x-axis and
positive along the _v-axis.
EXAMPLE 7 To get a feeling for what happens in the case of a degenerate
critical point (i.e., a critical point a such that det Hf(a) = 0), consider the three
functions
/(x,y) = x4 + x2 + y4,
g(x, y) = -x4 - x2
and
h(x, y) = x
x2 + y4.
We leave it to you to check that the origin (0, 0) is a degenerate critical point
of each of these functions. (In fact, the Hessians themselves look very similar.)
Since / is 0 at (0, 0) and strictly positive at all (x, y) ^ (0, 0), we see that /
has a strict minimum at the origin. Similar reasoning shows that g has a strict
maximum at the origin. For h, the situation is slightly more complicated. Along
the y-axis, we have h(0, y) = y4, which is zero at y = 0 (the origin) and strictly
positive everywhere else. Along the x-axis,
h(x,0) = x4
x2 = x2(x
1)(JC + 1).
For — 1 < x < 1 and x ^ 0, h(x, 0) < 0. We have the situation depicted in Fig-
ure 4. 19. Thus, every neighborhood of (0, 0) contains some points (x, y) where h
is positive and also some points where h is negative. Therefore, h has a saddle point
at the origin. The "moral of the story" is that a degenerate critical point can exhibit
any type of behavior, and more detailed consideration of the function itself, rather
than its Hessian, is necessary to understand its nature as a site of an extremum. ♦
Global Extrema on Compact Regions
Thus far our discussion has been limited to consideration of only local extrema.
We have said nothing about how to identify global extrema, because there really is
no general, effective method for looking at an arbitrary function and determining
whether and where it reaches an absolute maximum or minimum value. For the
purpose of applications, where finding an absolute maximum or minimum is
essential, such a state of affairs is indeed unfortunate. Nonetheless, we can say
something about global extrema for functions defined on a certain type of domain.
4.2 | Extrema of Functions 271
Figure 4.20 Compact regions.
DEFINITION 2.4 A subset X C R" is said to be compact if it is both closed
and bounded.
Recall that X is closed if it contains all the points that make up its boundary.
(See Definition 2.3 of §2.2.) To say that X is bounded means that there is some
(open or closed) ball B that contains it. (That is, X is bounded if there is
some positive number M such that ||x|| < M for all x e X.) Thus, compact sets
contain their boundaries (a consequence of being closed) and have only finite
extent (a consequence of being bounded). Some typical compact sets in R2 and
R3 are shown in Figure 4.20.
For our purposes the notion of compactness is of value because of the next
result, which we state without proof.
THEOREM 2.5 (Extreme value theorem) If X c R" is compact and /:
X -> R is continuous, then / must have both a global maximum and a global
minimum somewhere on X. That is, there must exist points amax and am;n in X
such that, for all x e X,
/(amin) < /(x) < /(amax).
We need the compactness hypothesis since a function defined over a noncom-
pact domain may increase or decrease without bound and hence, fail to have any
global extremum, as suggested by Figure 4.21. This is analogous to the situation
in one variable where a continuous function defined on an open interval may fail
to have any extrema, but one defined on a closed interval (which is a compact
subset of R) must attain both maximum and minimum values. (See Figure 4.22.)
In the one-variable case, extrema can occur either in the interior of the interval
or else at the endpoints. Therefore, you must compare the values of / at any
interior critical points with those at the endpoints to determine which is largest
and smallest. In the case of functions of n variables, we do something similar,
namely, compare the values of / at any critical points with values at any restricted
critical points that may occur along the boundary of the domain.
272 Chapter 4 I Maxima and Minima in Several Variables
x
Figure 4.21 A graph that lacks a
global minimum. Figure 4.22 The function depicted by the graph on the left has no global extrema — the
function is defined on the open interval (a, b). By contrast, the function defined on the
closed interval [a, b], and with the graph on the right, has both a global maximum and
minimum.
EXAMPLE 8 LetTiXcR2
y = 2
x = -l
x = 2
y = -l
R be given by
Figure 4.23 The domain of the
function T of Example 8.
T(x, y) = x2 -xy + y2+ 1,
where X is the closed square in Figure 4.23. (Note that X is compact.) Think of
the square as representing a fiat metal plate and the function T as the temperature
of the plate at each point. Finding the global extrema amounts to finding the
warmest and coldest points on the plate. According to Theorem 2.5, such points
must exist.
We need to find all possible critical points of T. Momentarily considering T
as a function on all of R2, we find the usual critical points by setting DT(x, y)
equal to 0. The result is the system of two equations
| 2x - y = 0
|-x + 2y = 0'
which has (0, 0) as its only solution. Whether it is a local maximum or minimum
is not important for now, because we seek global extrema. Because there is only
one critical point, at least one global extremum must occur along the boundary of
X (which consists of the four edges of the square). We now find all critical points
of the restriction of T to this boundary:
1. The bottom edge of X is the set
El={(x,y)\y = -l,-l<x <2}.
The restriction of T to E\ defines a new function f\ : [— 1 , 2] — > R given by
fx(x)= T(x,-\) = x2 +x + 2.
As f{(x) = 2x + 1, the function f\ has a critical point at x = — j. Thus,
we must examine the following points of X for possible extrema: (— \ , — l),
(— 1 , — 1), and (2, — 1). (The first point is the critical point of f\ , and the second
two are the vertices of X that lie on E\ .)
2. The top edge of X is given by
E2 = {(x,y) \ y = 2, -1 <x < 2}.
4.2 | Extrema of Functions 273
Consequently, we define fy. [— 1 , 2] -» R by
/2(jc) = T(jr, 2) = x2 - 2x + 5.
(/2 is the restriction of 71 to £2-) We calculate f^ipc) = 2x — 2, which implies
that x = 1 is a critical point of /2. Hence, we must consider (1, 2), (—1, 2),
and (2, 2) as possible sites for global extrema of T. (The points (—1,2) and
(2, 2) are the remaining two vertices of X.)
3. The left edge of X is
E3 = {(x,y)\x = -l,-\ < v<2}.
Therefore, we define fa : [— 1 , 2] -> R by
/3(y)=r(-l,y) = y2 + y + 2.
We have /3'(y) = 2y + 1, and so y = — I is the only critical point of fa. Thus
(— 1 , — 2) is a potential site of a global extremum. (We need not worry again
about the vertices (—1,-1) and (— 1 , 2).)
4. The right edge of X is
£4 = {(x,y)|x=2,-l <y<2}.
We define /4: [-1,2] -> Rby
f4(y) = 7X2, y) = y2 - 2y + 5.
We have /4'(y) = 2y — 2, and so y = 1 is the only critical point of /4. Hence,
we must include (2, 1) in our consideration.
Consequently, we have nine possible locations for global extrema, shown in
Figure 4.24. Now we need only to compare the actual values of T at these points
to see that (0, 0) is the coldest point on the plate and both (2, — 1) and (—1,2) are
the hottest points. ♦
If a function is defined over a noncompact region, there is no general result
like the extreme value theorem (Theorem 2.5) to guarantee existence of any global
extrema. However, ad hoc arguments frequently can be used to identify global
extrema.
(-1,2)
(1.2) (2,2)
(0,0)
(-1,-1)'
(-1,-1)/
H,-i)
(2,1)
(2,-1)
(*o0
T(x,y)
(0,0)
1
(-5,-1)
7
4
(-1,-1)
2
(2,-1)
8
(1,2)
4
(-1,2)
8
(2,2)
5
(-1,-1)
7
4
(2,1)
4
Figure 4.24 Possible global extrema for T .
274 Chapter 4 I Maxima and Minima in Several Variables
EXAMPLE 9 Consider the function f(x, y) = e1'3^ denned on all of R2
(so the domain is certainly not compact). Verifying that / has a unique critical
point at (0, 0) is straightforward. We leave it to you to check that the Hessian
criterion implies that / has a local maximum there. In any case, for all (x,y) 6 R2,
we have
1 - 3x2 - y2 < 1.
Therefore, because the exponential function is always increasing (i.e., if u\ < u%,
thene"1 < e"2),
As f(0, 0) = e, we see that / has a global maximum at (0, 0). ♦
WARNING It is tempting to assume that if a function has a unique critical point
that is a local extremum, then it must be a global extremum as well. Although
true for the case of a function of a single variable, it is not true for functions of
two or more variables. (See Exercise 52 for an example.)
Addendum: Proof of Theorem 2.3
Step 1. We show the following key property of a quadratic form Q, namely, that
if k e R, then
Q(Xh) = X2Q(h). (3)
This is straightforward to establish if we write Q in terms of its associated sym-
metric matrix B and use some of the properties of matrix arithmetic given in § 1 .6:
g(A.h) = (Xh)TB(Xh) = khTB(Xh) = X2hT Bh = A2g(h).
Step 2. We show that if B is the symmetric matrix associated to a positive
definite quadratic form Q, then there is a positive constant M such that
e(h) > Mpii2
for allh 6 R".
First, note that when h = 0, then Q(h) = <2(0) = 0 so the conclusion holds
trivially in this case.
Next, suppose that his a unit vector (i.e., ||h|| = 1). The (endpoints of the) set
of all unit vectors in R" is an (n — l)-dimensional sphere S, which is a compact
set. Hence, by the extreme value theorem (Theorem 2.5), the restriction of Q to
S must achieve a global minimum value M somewhere on S. Thus, Q(h) > M
for all h € S.
Finally, let h be any nonzero vector in R". Then its normalization h/||h|| is a
unit vector and so lies in S. Therefore, by the result of Step 1, we have
G(h)=e(l|l.|l1|||) = l|hfG(1|I)>IIN2«,
since h/||h|| is in S.
Step 3. Now we prove the theorem. By the second-order Taylor formula
Theorem 1.5 and formula (10) of §4.1, we have that, for the critical point a of /,
Af = f(x) - /(a) = i(x - a)r///(a)(x - a) + R2(x, a), (4)
where |i?2(x, a)|/||x — a||2 -> 0 as x -» a.
4.2 | Extrema of Functions 275
Suppose first that Hf(&) is positive definite. Then by Step 2 with h = x — a,
there must exist a constant M > 0 such that
i(x-a)r///(a)(x-a)> M||x-a||2. (5)
Because |i?2(x, a)|/||x — a||2 -> 0 as x -> a, there must be some 5 > 0 so that if
0 < ||x - a|| < <5, then |i?2(x, a)|/||x - a||2 < M, or, equivalently,
|i?2(x,a)| <M||x-a||2. (6)
Therefore, (4), (5), and (6) imply that, for 0 < ||x — a|| < <5,
A/>0
so that / has a (strict) local minimum at a.
If Hf(a) is negative definite, then consider g = —f. We see that a is also a
critical point of g and that Hg(a) = — Hf(a), so Hg(a) is positive definite. Hence,
the argument in the preceding paragraph shows that g has a local minimum at a,
so / has a local maximum at a.
Now suppose det Hf(a) 0, but that Hf(a) is neither positive nor negative
definite. Let xi be such that
i(x, - a)r///(a)(Xl - a) > 0
and X2 such that
i(x2 - a)r///(a)(x2 - a) < 0.
(Since det Hf(a) 0, such points must exist.) For i = 1,2 let
yt(0 = t(x - a) + a,
the vector parametric equation for the line through a and x,- . Applying formula
(4) with x = y,-(f), we see
A/ = f(Yi(t)) - f(a) = i(y,(f) - a)rH/(a)(y,(0 " a) + ^(y,(0, a)
= i(y(- 0 - a)rJf/(aXy,(0 - a) + ||y,(f) - a||2 g /
lly/(0-a||2
Note that y, (f) — a = f (x, — a). Therefore, using the property of quadratic forms
given in Step 1 and the fact that ||y,(?) - a||2 = ||?(x; - a)||2 = r2||x/ - a||2, we
have
/(y (0) - /(a)
(7)
= t2
i(x; - tfHfW* - a) + ||x, - a||2 f 2^(0' ^
llyi(0 - all2
Now note that, for i = 1 , the first term in the brackets in the right side of (7) is a
positive number P and, for i = 2, it is a negative number N. Set
M = min
|X! - a||2 ' ||x2-a||2
Because we know that \Rz(yi(t), a)|/||y(r) - a||2 -> 0 as t -+ 0, we can find
some S > 0 so that if 0 < t < S, then
1*2(7,(0, a)| <M
|y(0 -a||5
276 Chapter 4 [ Maxima and Minima in Several Variables
But this implies that, for 0 < t < S,
A/ = /(yi(O)-/(a)>0,
while
A/ = /(y2(0) " /(*) < 0.
Thus, / has a saddle point at x = a.
4.2 Exercises
1. Concerning the function f(x, y) = Ax + 6y — 12 —
2 2
x — y :
(a) There is a unique critical point. Find it.
(b) By considering the increment Af, determine
whether this critical point is a maximum, a mini-
mum, or a saddle point.
(c) Now use the Hessian criterion to determine the
nature of the critical point.
2. This problem concerns the function g(x, y) = x2 —
2y2 + 2x + 3.
(a) Find any critical points of g.
(b) Use the increment Ag to determine the nature of
the critical points of g.
(c) Use the Hessian criterion to determine the nature
of the critical points.
In Exercises 3-20, identify and determine the nature of the
critical points of the given functions.
3.
/(*. y) =
2xy - 2x2 - 5y2 + Ay - 3
4.
fix, y) =
ln(x2 + y2+ 1)
5.
fix, y) =
x2 + y3 — 6xy + 3x + 6y
6.
fix, y) =
y4 — 2xy2 + x3 — x
7.
fix, y) =
8 1
XV + - + -
x y
8.
fix, y) =
ex sin y
9.
fix, y) =
e-y(x2 - y2)
10.
fix, y) =
{X + y)(l - xy)
11.
fix, y) =
x — y — x y + y
12.
fix,y) =
e~x(x2 + 3y2)
13.
fix, y) =
2x — 3y + Inxy
14.
fix, y) =
cosx siny
15.
fix,y,z)
= x2 — xy + z2 — 2xz + 6z
16.
fix, y, z)
= (x2+2y2 + l)cosz
17.
fix, y, z)
= x2 + y2 + 2z2 + xz
18. f(x, y, z) = jc3 + xz2 - 3x2 + y2 + 2z2
1
19. f(x,y,z) = xy + xz + 2yz+ -
x
20. f(x,y,z) = ex(x2 - y2 - 2z2)
21. (a) Find all critical points of fix,y) =
2y3 - 3y2 - 36y + 2
1 + 3x2 '
(b) Identify any and all extrema of /.
22. (a) Under what conditions on the constant k will the
function
f{x, y) = kx2 — 2xy + ky2
have a nondegenerate local minimum at (0, 0)?
What about a local maximum?
(b) Under what conditions on the constant k will the
function
g(x, y, z) = kx2 + kxz — 2yz — y2 + -Z2
have a nondegenerate local maximum at (0, 0, 0)?
What about a nondegenerate local minimum?
23. (a) Consider the function f(x, y) = ax2 + by2,
where a and b are nonzero constants. Show that
the origin is the only critical point of /, and deter-
mine the nature of that critical point in terms of a
and b.
(b) Now consider the function f(x,y,z) = ax2 +
by2 + cz2, where a, b, and c are all nonzero. Show
that the origin in R3 is the only critical point of /,
and determine the nature of that critical point in
terms of a, b, and c.
(c) Finally, let fix\, x%, x„) = a\x\ + 02X2
+ ■ ■ ■ + a,,x2, where a; is a nonzero constant for
i = 1, 2, . . . , n. Show that the origin in R" is the
only critical point of /, and determine its nature.
Sometimes it can be difficult to determine the critical point of
a function f because the system of equations that arises from
setting V/ equal to zero may be very complicated to solve by
hand. For the functions given in Exercises 24—27, (a) use a
computer to assist you in identifying all the critical points of
the given function f , and (b) use a computer to construct the
4.2 I Exercises 277
Hessian matrix and determine the nature of the critical points
found in part (a).
^ 24. f(x, y) = y4 + x3 - 2xy2 - x
^> 25. f(x, y) = 2x3y - y2 - 3xy
^ 26. f(x, y, z) = yz- xyz - x2 - y2 - 2z2
27. f(x, y, z, u>) = yw — xyz — x2 — 2z2 + w2
28. Show that the largest rectangular box having a fixed
surface area must be a cube.
29. What point on the plane 3x — 4y — z = 24 is closest
to the origin?
30. Find the points on the surface xy + z2 = 4 that are
closest to the origin. Be sure to give a convincing ar-
gument that your answer is correct.
31 . Suppose that you are in charge of manufacturing two
types of television sets. The revenue function, in dol-
lars, is given by
R(x, y) = 8x + 6y - x2 - 2y2 + 2xy,
where x denotes the quantity of model X sets sold, and
y the quantity of model Y sets sold, both in units of
100. Determine the quantity of each type of set that
you should produce in order to maximize the resulting
revenue.
32. Find the absolute extrema of f(x, y) = x2 + xy +
y2 — 6y on the rectangle {(x, y) \ — 3 < x < 3, 0 <
y < 5}'.
33. Find the absolute maximum and minimum of
fix, y, z) = x2 + xz — y2 + 2z2 + xy + 5x
on the block {(x, y,z)\ - 5 < x < 0, 0 < y < 3,
0 < z < 2}.
34. A metal plate has the shape of the region x2 + y2 < 1 .
The plate is heated so that the temperature at any point
(x , y) on it is indicated by
T(x,y) = 2x2 + y2-y + 3.
Find the hottest and coldest points on the plate and the
temperature at each of these points. (Hint: Parametrize
the boundary of the plate in order to find any critical
points there.)
35. Find the (absolute) maximum and minimum values of
f(x, y) = sinx cosy on the square R = {(x, y) | 0 <
x < 2it, 0 < y < 2jt}.
36. Find the absolute extrema of f(x, y) = 2 cosx +
3 siny on the rectangle {(x, y) \ 0 < x < 4, 0 <
y <3}.
37. Determine the absolute minimum and maximum
values of the function f(x, y) = 2x2 — 2xy + y2
— y + 3 on the closed triangular region with vertices
(0, 0), (2, 0), and (0, 2).
38. Determine the absolute minimum and maximum val-
ues of the function f(x, y) = x2y on the elliptical re-
gion D = {(x, y) | 3x2 + 4y2 < 12}.
39. Find the absolute extrema of f(x, y, z) =
e!-x2_y2+2y_z2_4z Qn ^ ball {(^ _^ z) | x2 + y2 - 2y
+ z2 + 4z < 0}.
Each of the functions in Exercises 40-45 has a critical point
at the origin. For each function, (a) check that the Hessian
fails to provide any information about the nature of the critical
point at the origin, and (b) find another way to determine if the
function has a maximum, minimum, or neither at the origin.
40. f(x,y) = x2y2
41. f(x,y) = 4-3x2y2
42. f(x,y) = x3y3
43. f(x, y, z) = x2y3z4
44. f(x, y, z) = x2y2z4
45. f(x,y,z) = 2-x4y4 -z4
In Exercises 46—48, (a) find all critical points of the given
function f and identify their nature as local extrema and (b)
determine, with explanation, any global extrema of f.
46. f(x,y) = ex2+5>2
47. f(x,y, z) = e2-*2-2}2-3'-4
48. f(x, y) = x3 + y3 - 3xy + 7
49. Determine the global extrema, if any, of
f(x, y) = xy + 2y — lnx — 2 lny,
where x, y > 0.
50. Find all local and global extrema of the function
f{x, y, z) = x3 + 3x2 + e>,2+l +z2- 3xz.
51. Let f(x, y) = 3 - [(x - l)(y - 2)]2/3.
(a) Determine all critical points of /.
(b) Identify all extrema of /.
52. (a) Suppose /: R — > R is a differentiate function of
a single variable. Show that if / has a unique crit-
ical point at xo that is the site of a strict local ex-
tremum of /, then / must attain a global extremum
at xo.
(b) Let fix, y) = 3yex - e3x - y3. Verify that / has
a unique critical point and that / attains a local
maximum there. However, show that / does not
have a global maximum by considering how / be-
haves along the y-axis. Hence, the result of part
(a) does not carry over to functions of more than
one variable.
278 Chapter 4 I Maxima and Minima in Several Variables
53. (a) Let / be a continuous function of one variable.
Show that if / has two local maxima, then / must
also have a local minimum.
(b) The analogue of part (a) does not necessarily hold
for continuous functions of more than one variable,
as we now see. Consider the function
f(x,y) = 2-(xy2
D2-(v2-l)2
Show that / has just two critical points — and that
both of them are local maxima.
(c) Use a computer to graph the function / in part (b).
4.3 Lagrange Multipliers
Constrained Extrema
Frequently, when working with applications of calculus, you will find that you
do not need simply to maximize or minimize a function but that you must do so
subject to one or more additional constraints that depend on the specifics of the
situation. The following example is a typical situation:
EXAMPLE 1 An open rectangular box is to be manufactured having a (fixed)
volume of 4 ft3 . What dimensions should the box have so as to minimize the
amount of material used to make it?
We'll let the three dimensions of the box be independent variables x, y, and
z, shown in Figure 4.25. To determine how to use as little material as possible,
we need to minimize the surface area function A given by
A(x, y, z) = 2xy + 2yz + xz
front and back sides bottom only
For x, y, z > 0, this function has neither minimum nor maximum. However, we
have not yet made use of the fact that the volume is to be maintained at a constant
4 ft3 . This fact provides a constraint equation,
V(x, y, z) = xyz = 4.
The constraint is absolutely essential if we are to solve the problem. In particular,
the constraint enables us to solve for z in terms of x and y:
4
xy'
We can thus create a new area function of only two variables:
/ 4
a(x, y) = A j x, y, —
V xy
8 4
= 2xy+ - + -.
x y
Now we can find the critical points of a by setting Da equal to 0:
v
Figure 4.25 The open box of
Example 1.
4.3 | Lagrange Multipliers
The first equation implies
so that the second equation becomes
2x - 4
16
or, equivalently,
x 1 - -x* = 0
The solutions to this equation are x = 0 (which we reject) and x = 2. Thus, the
critical point of a of interest is (2, 1), and the constrained critical point of the
original function A is (2, 1,2).
We can use the Hessian criterion to check that x = 2, y = 1 yields a local
minimum of a :
Ha(x, y) =
16/x3
2
2
8/y3
so Ha(2, 1) =
The sequence of minors is 2, 12 so we conclude that (2, 1) does yield a local
minimum of a. Because a(x, y) — >■ oo as either jc — > 0+, y — > 0+, je — > oo, or
y -> oo, we conclude that the critical point must yield a global minimum as well.
Thus, the solution to the original question is to make the box with a square base
of side 2 ft and a height of 1 ft. ♦
The abstract setting for the situation discussed in Example 1 is to find max-
ima or minima of a function f(x\, xi, ■ ■ ■ , xn) subject to the constraint that
g{x\, X2, ■ ■ ■ , xn) = c for some function g and constant c. (In Example 1, the
function / is A(x, y, z), and the constraint is xyz = 4.) One method for finding
constrained critical points is used implicitly in Example 1: Use the constraint
equation g(x) = c to solve for one of the variables in terms of the others. Then
substitute for this variable in the expression for /(x), thereby creating a new
function of one fewer variables. This new function can then be maximized or
minimized using the techniques of §4.2. In theory, this is an entirely appropriate
way to approach such problems, but in practice there is one major drawback: It
may be impossible to solve explicitly for any one of the variables in terms of the
others. For example, you might wish to maximize
f(x,y,z) = x2 + 3y2 + y2z4
subject to
g(x,y,z) = ex
x5y2z + cos
= 2.
There is no means of isolating any of x, y, or z on one side of the constraint
equation, and so it is impossible for us to proceed any further along the lines of
Example 1.
280 Chapter 4 | Maxima and Minima in Several Variables
The Lagrange Multiplier
The previous discussion points to the desirability of having another method for
solving constrained optimization problems. The key to such an alternative method
is the following theorem:
THEOREM 3.1 Let X be open in R" and f,g:X^Rbe functions of class C1 .
Let S = {x e X \ g(x) = c} denote the level set of g at height c. Then if / \s (the
restriction of / to S) has an externum at a point x0eS such that Vg(x0) 0,
there must be some scalar X such that
V/(x0) = AVg(x0).
The conclusion of Theorem 3.1 implies that to find possible sites for extrema
of / subject to the constraint that g(x) = c, we can proceed in the following
manner:
1. Form the vector equation V/(x) = XVg(x).
2. Solve the system
V/(x) = AVg(x)
g(X) = c
for x and X. When expanded this is actually a system of n + 1 equations in
n + 1 unknowns x\, x2, ■ ■ ■ , xn, X, namely,
fXl(xi,x2, ■ ■ .,x„) = kgXl(xi,x2, ■ ■ -,x„)
fX2(xi,x2, ...,xn) = XgX2(xi,x2, ■ ■ ■ , xn)
•
fx„(xi >x2,...,xn) = XgXiXxi ,x2, xn)
g(xux2, . . . ,x„) = c
The solutions for x = (x\, x2, . . . , xn) in the system above, along with any
other points x satisfying the constraint g(x) = c and such that V/ is unde-
fined or Vg vanishes or is undefined, are the candidates for extrema for the
problem.
3. Determine the nature of / (as maximum, minimum, or neither) at the critical
points found in Step 2.
The scalar A appearing in Theorem 3. 1 is called a Lagrange multiplier, after
the Italian-born French mathematician Joseph-Louis Lagrange (1736-1813) who
first developed this method for solving constrained optimization problems. In
practice, Step 2 can involve some algebra, so it is important to keep your work
organized. (Alternatively, you can use a computer to solve the system.) In fact,
since the Lagrange multiplier X is usually not of primary interest, you can avoid
solving for it explicitly, thereby reducing the algebra and arithmetic somewhat.
Determining the nature of a constrained critical point (Step 3) can be a tricky
business. We'll have more to say about that issue in the examples and discussions
that follow.
EXAMPLE 2 Let us use the method of Lagrange multipliers to identify the
critical point found in Example 1 . Thus, we wish to find the minimum of
A(x, y, z) = 2xy + 2yz + xz
4.3 | Lagrange Multipliers 281
subject to the constraint
V(x, y, z) = xyz = 4.
Theorem 3.1 suggests that we form the equation
VA(x,y,z) = XVV(x,y,z).
This relation of gradients coupled with the constraint equation gives rise to the
system
2y + z = kyz
2x + 2z = ~kxz
2y + x = kxy
xyz = 4
Since A is not essential for our final solution, we can eliminate it by means of any
of the first three equations. Hence,
_ 2y + z _2x +2z _2y + x
yz xz xy
Simplifying, this implies that
2 1 2 2 2 1
- + - = - + - = - + -.
z y z x x y
The first equality yields
— = — or x = 2y,
y x
while the second equality implies that
2 1
- = - or z = 2y.
z y
Substituting these relations into the constraint equation xyz = 4 yields
(2y)(y)(2y) = 4,
so that we find that the only solution isy = l,je=z = 2, which agrees with our
work in Example 1. (Note that V V = 0 only along the coordinate axes, and such
points do not satisfy the constraint V(x, y, z) = 4.) ♦
An interesting consequence of Theorem 3. 1 is this: By Theorem 6.4 of Chap-
ter 2, we know that the gradient Vg, when nonzero, is perpendicular to the level
sets of g. Thus, the equation V/ = XV g gives the condition for the normal vector
to a level set of / to be parallel to that of a level set of g. Hence, for a point xo
to be the site of an extremum of /' on the level set S = {x | g(x) = c], where
Vg(x0) ^ 0, we must have that the level set R of / that contains x0 is tangent to
S at xo.
EXAMPLE 3 Consider the problem of finding the extrema of f(x, y) = x2/4
+ y2 subject to the condition that x2 + y2 = 1. We let g(x, y) = x2 + y2, and
so the Lagrange multiplier equation V f(x, y) = XVg(x, y), along with the
282 Chapter 4 | Maxima and Minima in Several Variables
constraint equation, yields the system
x
Figure 4.26 The level sets of the
function f(x, y) = x2/4 + y2
define a family of ellipses. The
extrema of / subject to the
constraint that x2 + y2 = 1 (i.e.,
that lie on the unit circle) occur at
points where an ellipse of the
family is tangent to the unit circle.
= 2Xx
2y = 2Xy ■
x2 + y2 = 1
(There are no points simultaneously satisfying g(x, y) = 1 and Vg(x,y) =
(0, 0).) The first equation of this system implies that either x = 0 or X = \.
If x = 0, then the second two equations, taken together, imply that y = ±1 and
X = 1 . If X = \, then the second two equations imply y = 0 and x — ±1 . There-
fore, there are four constrained critical points: (0, ±1), corresponding to X = 1,
and (±1, 0), corresponding to X = |.
We can understand the nature of these critical points by using geometry and
the preceding remarks. The collection of level sets of the function / is the family
of ellipses x2/4 + y2 = k whose major and minor axes lie along the x- and y-
axes, respectively. In fact, the value f(x, y) = x2 /A + y2 = k is the square of
the length of the semiminor axis of the ellipse x2/4 + y2 = k. The optimization
problem then is to find those points on the unit circle x2 + y2 = 1 that, when
considered as points in the family of ellipses, minimize and maximize the length
of the minor axis. When we view the problem in this way, we see that such points
must occur where the circle is tangent to one of the ellipses in the family. A sketch
shows that constrained minima of / occur at (± 1 , 0) and constrained maxima at
(0, ± 1 ). In this case, the Lagrange multiplier X represents the square of the length
of the semiminor axis. (See Figure 4.26.) ♦
EXAMPLE 4 Consider the problem of determining the extrema of f(x, y) =
2x + y subject to the constraint that *Jx + Jy = 3. We let g(x, y) = *Jx + ^/y,
so that the Lagrange multiplier equation V f(x, y) = XVg(x, y), along with the
constraint equation, yields the system
X
2jx
X
1 = ■
yfx + s/y = 3
The first two equations of this system imply that X = 4^/x = 2Jy so that Jy =
2yfx. Using this in the last equation, we find that 3^/x = 3 and hence, x — 1.
Thus, the system of equations above yields the unique solution (1,4).
Since the constraint defines a closed bounded curve segment, the extreme
value theorem (Theorem 2.5) applies to guarantee that / must attain both a
global maximum and a global minimum on this segment. However, the Lagrange
multiplier method has provided us with just a single critical point. But note that the
points (9, 0) and (0, 9) satisfy the constraint *Jx + ^/y = 3; they are both points
where Vg is undefined. Moreover, we have /(l, 4) = 2, while f(9, 0) = 18 and
f(0, 9) = 9. Evidently then, the minimum of / occurs at (1 , 4) and the maximum
at (9,0).
We can understand the geometry of the situation in the following manner. The
collection of level sets of the function / is the family of parallel lines 2x + y = k.
Note that the height k of each level set is just the y-intercept of the corresponding
line in the family. Thus, the problem we are considering is to find the largest and
4.3 | Lagrange Multipliers 283
y
Figure 4.27 The level sets of the function f(x, y) = 2x + y define a family
of lines. The minimum of / subject to the constraint that *Jx + *Jy = 3
occurs at a point where one of the lines is tangent to the constraint curve and
the maximum at one of the endpoints of the curve.
smallest y -intercepts of any line in the family that meets the curve *Jx + Jy = 3.
These extreme values of k occur either when one of the lines is tangent to the
constraint curve or at an endpoint of the curve. (See Figure 4.27.)
This example illustrates the importance of locating all the points where ex-
trema may occur by considering places where V/ or Vg is undefined (or where
Vg = 0) as well as the solutions to the system of equations determined using
Lagrange multipliers. ♦
Figure 4.28 The gradient
Vg(xo) is perpendicular to
S = {x | g(x) = c}, hence, to the
tangent vector at xo to any curve
x(r) lying in S and passing through
xo. If / has an extremum at xo,
then the restriction of / to the
curve also has an extremum at xq.
Sketch of a proof of Theorem 3.1 We present the key ideas of the proof, which
are geometric in nature. Try to visualize the situation for the case n = 3, where
the constraint equation g(x, y, z) = c defines a surface S inR3. (See Figure 4.28.)
In general, if S is defined as {x | g(x) = c) with Vg(xo) ^ 0, then (at least locally
near xo) S is a hypersurface in R". The proof that this is the case involves the
implicit function theorem (Theorem 6.5 in §2.6), and this is why our proof here
is just a sketch.
Thus, suppose that x0 is an extremum of / restricted to S. We consider a
further restriction of / — to a curve lying in S and passing through x0. This will
enable us to use results from one-variable calculus. The notation and analytic
particulars are as follows: Let x: / c R —> S C R3 be a C1 path lying in S with
\(to) = xo for some to 6 / . Then the restriction of / to x is given by the function
F, where
F(f) = /(x(0).
Because xo is an extremum of / on S, it must also be an extremum on x. Conse-
quently, we must have F'(to) = 0, and the chain rule implies that
= ^/(x(0)|r=ro = V/(x(f0)).x'(fo) = V/(x0)-x'(f0).
Thus, V/(xo) is perpendicular to any curve in S passing through x0; that is,
V /(xo) is normal to S at xo. We've seen previously in §2.6 that the gradient
Vg(xo) is also normal to S at xq. Since the normal direction to the level set S is
284 Chapter 4 [ Maxima and Minima in Several Variables
uniquely determined and Vg(xo) 7^ 0, we must conclude that V /(xo) and Vg(xo)
are parallel vectors. Therefore,
V/(x0) = AVg(xo)
for some scalar X e R, as desired. ■
The Case of More than One Constraint
It is natural to generalize the situation of finding extrema of a function / subject
to a single constraint equation to that of finding extrema subject to several con-
straints. In other words, we may wish to maximize or minimize / subject to k
simultaneous conditions of the form
gi(x) = c\
g2(x) = C2
gk(x) = Ck
The result that generalizes Theorem 3.1 is as follows:
THEOREM 3.2 Let X be open in R" and let /, gu . . . , gk: X C R" -> R be C1
functions, where k < n.LetS = [x e X \ g\(x) = c\, .. ., gfc(x) = ck}.lf f\shas
an extremum at a point xo, where Vgi(xo), . . . , Vgt(xo) are linearly independent
vectors, then there must exist scalars X\, ... ,Xk such that
V/(xo) = AjVgKxo) + A2Vg2(x0) + ■ ■ ■ + XkVgk(x0).
(Note: k vectors vi , . . . , in R" are said to be linearly independent if the only
way to satisfy a\ vi + ■ ■ ■ + akvk = 0 for scalars a\, . . . , ak is if a\ = a% = ■ ■ ■ =
ak = 0.)
Idea of proof First, note that S is the intersection of the k hypersurfaces Si , . . . ,
S/c, where Sj = {x 6 R" | g/(x) = cj}. Therefore, any vector tangent to S must
also be tangent to each of these hypersurfaces, and so, by Theorem 6.4 of
Chapter 2, perpendicular to each of the Vg; 's. Given these remarks, the main
ideas of the proof of Theorem 3.1 can be readily adapted to provide a proof of
Theorem 3.2.
Therefore, we let xo e 5 be an extremum of / restricted to S and consider
the one-variable function obtained by further restricting / to a curve in S through
xo. Thus, let x: / -> S C R" be a C1 curve in S with x(f0) = xo for some to £ I.
Then, as in the proof of Theorem 3.1, we define F by
F(t) = /(x(0).
It follows, since x0 is assumed to be a constrained extremum, that
F'(t0) = 0.
The chain rule then tells us that
0 = F'(to) = V/(x(*0)) • x'(to) = V/(xo) • x'(to).
That is, V /(xo) is perpendicular to all vectors tangent to S at xo. Therefore, it can
be shown that V/(xq) is in the ^-dimensional plane spanned by the normal vectors
4.3 | Lagrange Multipliers 285
Plane spanned by
Vg^f-^i^o) and
Figure 4.29 Illustration of the
proof of Theorem 3.2. The
constraints gi(x) = c\ and
g2(x) = C2 are the surfaces S\ and
S2 • Any extremum of / must occur
at points where V/ is in the plane
spanned by Vgi and Vg2.
to the individual hypersurfaces S\, . .. , Sk whose intersection is S. It follows (via
a little more linear algebra) that there must be scalars ki, . . . , kk such that
V/(xo) = hVgi(xo) + A2Vg2(x0) +
■kkWgk(x0).
A suggestion of the geometry of this proof is provided by Figure 4.29 (where
k = 2 and n = 3). ■
EXAMPLE 5 Suppose the cone z2 = x2 + y2 is sliced by the plane z = x +
y + 2 so that a conic section C is created. We use Lagrange multipliers to find
the points on C that are nearest to and farthest from the origin in R3 .
The problem is to find the minimum and maximum distances from (0, 0, 0)
of points (x, y, z) on C. For algebraic simplicity, we look at the square of the
distance rather than the actual distance. Thus, we desire to find the extrema of
f(x, y, Z) = x2 + y2 + z2
(the square of the distance from the origin to (x, y, z)) subject to the constraints
gi(x, y, z) = x2 + y2 - z2 = 0
g2(x, y,z) = x + y - z = -2
Note that
Vgi(x, y, z) = (2x, 2y,
-2z) and Vg2(x,y,z) = (1,1,-1).
These vectors are linearly dependent only when x = y = z- However, no point of
the form (x, x, x) simultaneously satisfies g\ = 0 and g2 = —2. Hence, Vgi and
Vg2 are linearly independent at all points that satisfy the two constraints. There-
fore, by Theorem 3.2, we know that any constrained critical points (x0, y0, Zo)
must satisfy
V/(*o, yo, z0) = kiVgi(x0, y0, zo) + k2Vg2(x0, yo, zo),
as well as the two constraint equations. Thus, we must solve the system
2x = 2k\x + k2
2y = 2kiy + k2
2z = -2k\z — k2 .
x2 + y2-z2 = 0
x + y - z = -2
Eliminating k2 from the first two equations yields
k2 = 2x — 2k\x =2y — 2A.iv,
which implies that
Therefore, either
2(x - y)(l - M) = 0.
x = y or A.1 = 1.
The condition ki = 1 implies immediately k2 = 0, and the third equation of the
system becomes 2z = —2z, so z must equal 0. If z = 0, then x and y must be
286 Chapter 4 | Maxima and Minima in Several Variables
zero by the fourth equation. However, (0, 0, 0) is not a point on the plane z =
x + y + 2. Thus, the condition X \ = 1 leads to no critical point. On the other hand,
if x = y, then the constraint equations (the last two in the original system of five)
become
2x2 -z2 = 0
2x — z = —2
Substituting z = 2x + 2 yields
2x2 - (2x + 2)2 = 0,
equivalent to
2x2 + 8x + 4 = 0,
whose solutions are x = — 2 ± V2- Therefore, there are two constrained critical
points
ai = (-2 + 72, -2 + 4l, -2 + 2V2)
and
a2 = (-2 - V2, -2 - V2, -2 - 2V2) .
We can check that
/(ai) = 24 - 16 V2, /(a2) = 24 + 16^2,
so it seems that ai must be the point on C lying nearest the origin, and a2 must be
the point that lies farthest. However, we don't know a priori if there is a farthest
point. If the conic section C is a hyperbola or a parabola, then there is no point
that is farthest from the origin. To understand what kind of curve C is, note
that ai has positive z-coordinate and a2 has negative z-coordinate. Therefore, the
plane z = x + y + 2 intersects both nappes of the cone z2 = x2 + y2. The only
conic section that intersects both nappes of a cone is a hyperbola. Hence, C is a
hyperbola, and we see that the point ai is indeed the point nearest the origin, but
the point a2 is not the farthest point. Instead a2 is the point nearest the origin on
the branch of the hyperbola not containing aj. That is, local constrained minima
occur at both ai and a2, but only &\ is the site of the global minimum. (See
Figure 4.30.) ♦
A Hessian Criterion for Constrained Extrema (optional) —
As Example 5 indicates, it is often possible to determine the nature of a critical
point (constrained or unconstrained) from considerations particular to the prob-
lem at hand. Sometimes this is not difficult to do in practice and can provide
useful insight into the problem. Nonetheless, occasionally it is advantageous to
have a more automatic means of discerning the nature of a constrained critical
point. We therefore present a Hessian criterion for constrained critical points.
Like the one in the unconstrained case, this criterion only determines the local
nature of a critical point. It does not provide information about global constrained
extrema.2
Figure 4.30 The
point ai is the point
on the hyperbola
closest to the origin.
The point a2 is the
point on the lower
branch of the
hyperbola closest
to the origin.
2 We invite the reader to consult D. Spring, Amer. Math. Monthly, 92 (1985), no. 9, 631-643 for a more
complete discussion.
4.3 | Lagrange Multipliers 287
In general, the context for the Hessian criterion is this: We seek extrema of a
function /:XcR"-^R subject to the k constraints
gi(xi,x2, . . .,x„) = a
g2(X\,X2, X„) = C2
gk(X\,X2,
i Xfi) — Ck
We assume that f,gi,...,gk are all of class C2, and assume, for simplicity, that
/ and the gj 's all have the same domain X. Finally, we assume that Vg\ , . . . , Vgk
are linearly independent at the constrained critical point a. Then, by Theorem 3.2,
any constrained extremum a must satisfy
V/(a) = A.i Vgi(a) + A2Vg2(a) + ■ ■ • + A.tVgt(a)
for some scalars Ai , . . . , A*. We can consider a constrained critical point to be a
pair of vectors
(A;a) = (Ai, ...,Xk;au ...,an)
satisfying the aforementioned equation. In fact, we can check that (A; a) is an
unconstrained critical point of the so-called Lagrangian function L defined
by
L(h, ...,lk;xi,...,x„)= f{x\, . . . ,xn) - y^Ji(gj(xi, ...,xn)- a).
The Hessian criterion comes from considering the Hessian of L at the critical
point (A; a). Before we give the criterion, we note the following fact from linear
algebra: Since Vgi(a), . . . , Vgj(a) are assumed to be linearly independent, the
derivative matrix of g = (g\ , . . . , gk) at a,
Dg(a) =
dx\
(a)
dx
(a)
9gi
dxn
dgk
dxn
(a)
(a)
hasa^: x k submatrix( obtained by deleting n — k columns ofDg(a)) with nonzero
determinant. By relabeling the variables if necessary, we will assume that
3jc
(a)
dxk
(a)
det
^0
dx
(a)
dxk
(a)
(i.e., that we may delete the last n — k columns).
Chapter 4 I Maxi ma and M
inima in Several Variables
Second derivative test for constrained local extrema. Given a constrained
critical point a of / subject to the conditions g\(x) = c\, g2(x) = c2, ■ . . ,
gk(x) = Ck, consider the matrix
HL(X; a) =
0
dx
(a)
dxn
(a)
0
0
dgk
dx\
dgk
dxn
dx\
(a)
dxn
(a)
(a)
(a)
dx\
(a)
dx„
(a)
where
htj =
d2f
(a) - h - - (a) - X2 (a)
OXj dXi
— —(a).
dxjdxj dxjdxi dxjdxt dxjdxt
(Note that HL(k; a) is an (« + /:) x + /c) matrix.) By relabeling the vari-
ables as necessary, assume that
det
3jc
(a)
3x
(a)
3*<t
3xi.
(a)
(a)
^0.
As in the unconstrained case, let be the upper leftmost j x j subma-
trix of HL(X, a). For j = 1, 2, . . . , k + «, let <2;- = det and calculate the
following sequence of « — k numbers:
(-1)^+2,..., (1)
Note that, if k > 1, the sequence in (1) is not the complete sequence of prin-
cipal minors of HL(X, a). Assume *4+n = det HL(X, a) / 0. The numerical
test is as follows:
1. If the sequence in (1) consists entirely of positive numbers, then / has a
local minimum at a subject to the constraints
gi(x) = ci, g2(x) = c2,
gk(*) = ck.
2. If the sequence in (1) begins with a negative number and thereafter alter-
nates in sign, then / has a local maximum at a subject to the constraints
gl(x) = Ci, g2(x) = C2,
gk(x) = Ck.
3. If neither case 1 nor case 2 holds, then / has a constrained saddle point
at a.
In the event that det HL(X, a) = 0, the constrained critical point a is degen-
erate, and we must use another method to determine whether or not it is the
site of an extremum.
4.3 | Lagrange Multipliers 289
Finally, in the case of no constraint equations g, (x) = c, (i.e., k = 0), the
preceding criterion becomes the usual Hessian test for a function / of n variables.
EXAMPLE 6 In Example 1, we found the minimum of the area function
A(x, v, z) = 2xy + 2yz + xz
of an open rectangular box subject to the condition
V(x, y, z) = xyz = 4.
Using Lagrange multipliers, we found that the only constrained critical point was
(2, 1 , 2). The value of the multiplier A. corresponding to this point is 2. To use the
Hessian criterion to check that (2, 1, 2) really does yield a local minimum, we
construct the Lagrangian function
L(l;x, y, z) = A(x, y, z) - l(V(x, y, z) - 4)
= 2xy + 2yz + xz - l{xyz — 4).
Then
HL(l;x,y,z) =
0
-yz
-xz
-xy
-yz
0
2 — Ix
l-ly
-xz
2-lz
0
2 — Ix
-xy
l-ly
2 — Ix
0
At the constrained critical point (2; 2, 1 , 2), we have
HL{2;2, 1,2)
The sequence of determinants to consider is
(-l)1det//2(1)+1 =
(-iydet//4
0
-2
-4 "
det
-2
0
-2
= 32,
-4
-2
0
- 0
-2
-4
-2 "
det
-2
-4
0
-2
-2
0
-1
-2
_ -2
-1
-2
0
48.
Since these numbers are both positive, we see that (2, 1,2) indeed minimizes the
area of the box subject to the constant volume constraint. ♦
EXAMPLE 7 In Example 5, we found points on the conic section C denned
by equations
\gl(x,y,z) = x2 + y2-z2 = 0
I gi(x, y, z) = x + y - z = -2
that are (constrained) critical points of the "distance" function
f(x,y,z) = x2 + y2 + z2.
To apply the Hessian criterion in this case, we construct the Lagrangian function
z2) - m{x + y - z + 2).
L(l,m;x, y, z) = x2 + y2 + z2
l(x2 + y2
290 Chapter 4 I Maxima and Minima in Several Variables
The critical points of L, found by setting DL(l, m;x,y, z) equal to 0, are
(A.i;ai) = (-3 + 2V2,
-24 +
16^2 ;
-2 + wl,
-2 + wl, -
-2 +
and
(A2;a2) = (-3-2V2,-
-24-
16V2;-
-2 - y/2,
-2-V2, -
- 2 -
The Hessian of L is
0
0
—2x
-2y
2z
0
0
-1
-1
1
HL(l, m;x, y, z) =
— 2x
-1
2-2/
0
0
-2y
-1
0
2-2/
0
2z
1
0
0 2
+ 21
2V2).
After we evaluate this matrix at each of the critical points, we need to compute
(-l)2det//2(2)+1 =det//5.
We leave it to you to check that for (k\; ai) this determinant is 128 — 64 V2 f*s
37.49, and for (A2; a2) it is 128 + 64V2 & 218.51. Since both numbers are pos-
itive, the points (—2 ± V2> —2 ± -Jl, —2 ± 2%/2) are both sites of local min-
ima. By comparing the values of / at these two points, we see that (—2 + \/2,
—2 + V2, —2 + 2*Jl) must be the global minimum. ♦
4.3 Exercises
1 . In this problem, find the point on the plane 2x — 3 y —
z = 4 that is closest to the origin in two ways:
(a) by using the methods in §4.2 (i.e., by finding the
minimum value of an appropriate function of two
variables);
(b) by using a Lagrange multiplier.
In Exercises 2-12, use Lagrange multipliers to identify the crit-
ical points of f subject to the given constraints.
2. f(x, y) = y, 2x2 + y2 = 4
3. f(x, y) = 5x + 2y, 5x2 + 2y2 = 14
4. f(x, y) = xy, 2x — 3y = 6
5. f(x, y, z) = xyz, 2x + 3 y + z = 6
6. f(x, y,z) = x2 + y2 + z2, X + y - z = 1
7. f(x,y,z) = 3-x2-2y2-z2, 2x + y + z = 2
8. f(x, y, z) = x6 + y6 + z6, x2 + y2 + z2 = 6
9. f(x, y, z) = 2x + y2-z2, x - 2y = 0, x + z = 0
10. f(x, y, z) = 2x + y2 + 2z, x2 - y2 = 1, x + y +
z = 2
11. f(x, y, z) = xy + yz, x2 + y2 = 1, yz = 1
12. f(x, y,z) = x + y + z, y2 - x2 = 1, x + 2z=l
13. (a) Find the critical points of f(x, y) = x2 + y sub-
ject to x2 + 2y2 = 1.
(b) Use the Hessian criterion to determine the nature
of the critical point.
14. (a) Find any critical points of f(x, y, z, w) = x2 +
y2 + z2 + w2 subject to 2x + y + z = 1 , x — 2z —
w = —2, 3x + y + 2w = — 1.
(b) Use the Hessian criterion to determine the nature
of the critical point. (Note: You may wish to use a
computer algebra system for the calculations.)
Just as sometimes is the case when finding ordinary (i.e., un-
constrained) critical points of functions, it can be difficult to
solve a Lagrange multiplier problem because the system of
equations that results may be prohibitively difficult to solve by
hand. In Exercises 15—19, use a computer algebra system to
find the critical points of the given function f subject to the
constraints indicated. (Note: You may find it helpful to provide
numerical approximations in some cases.)
^ 15. f(x, y, z) = 3xy - 4z, 3x + y - 2xz = 1
^ 16. f(x, y, z) = 3xy - 4yz + 5xz, 3x + y + 2z= 12,
2x - 3y + 5z = 0
4.3 I Exercises 291
17. f(x, y, z) = y3 + 2xyz - x2, x2 + y2 + z2 = 1
18. f(x, y, z) = x2 + y2 - xz2 , xy + z2 = 1
19. f(x, y, z, w) = x2 + y2 + z2 + w2, x2 + y2 = 1,
x + y + z + w = \,x — y + z — w = 0
20. Consider the problem of determining the extreme val-
ues of the function f(x, y) = x3 + 3y2 subject to the
constraint that xy = —A.
(a) Use a Lagrange multiplier to find the critical points
of / that satisfy the constraint.
(b) Give an analytic argument to determine if the criti-
cal points you found in part (a) yield (constrained)
maxima or minima of /.
(c) Use a computer to plot, on a single set of axes, sev-
eral level curves of / together with the constraint
curve xy = —A. Use your plot to give a geometric
justification for your answers in parts (a) and (b).
21. Find three positive numbers whose sum is 18 and
whose product is as large as possible.
22. Find the maximum and minimum values of
f(x, y, z) = x + y — z on the sphere x2 + y2 + z2 =
8 1 . Explain how you know that there must be both a
maximum and a minimum attained.
23. Find the maximum and minimum values of f(x, y) =
x2 + xy + v2 on the closed disk D = {(x, y) | x2 + y2
<4}. '
24. You are sending a birthday present to your calculus in-
structor. Fly-By-Night Delivery Service insists that any
package it ships be such that the sum of the length plus
the girth be at most 1 08 in. (The girth is the perimeter of
the cross section perpendicular to the length axis — see
Figure 4.31.) What are the dimensions of the largest
present you can send?
Figure 4.31 Diagram for
Exercise 24.
25. A cylindrical metal can is to be manufactured from
a fixed amount of sheet metal. Use the method of
Lagrange multipliers to determine the ratio between
the dimensions of the can with the largest capacity.
26. An industrious farmer is designing a silo to hold her
900tt ft3 supply of grain. The silo is to be cylindrical
in shape with a hemispherical roof. (See Figure 4.32.)
Suppose that it costs five times as much (per square
foot of sheet metal used) to fashion the roof of the silo
as it does to make the circular floor and twice as much
to make the cylindrical walls as the floor. If you were
to act as consultant for this project, what dimensions
would you recommend so that the total cost would be
a minimum? On what do you base your recommen-
dation? (Assume that the entire silo can be filled with
grain.)
Figure 4.32 The grain silo
of Exercise 26.
27. You are in charge of erecting a space probe on the
newly discovered planet Nilrebo. To minimize interfer-
ence to the probe's sensors, you must place the probe
where the magnetic field of the planet is weakest. Nil-
rebo is perfectly spherical with a radius of 3 (where
the units are thousands of miles). Based on a coordi-
nate system whose origin is at the center of Nilrebo,
the strength of the magnetic field in space is given by
the function M(x, y, z) = xz — y2 + 3x + 3. Where
should you locate the probe?
28. Heron's formula for the area of a triangle whose sides
have lengths x, y, and z is
Area = ^s(s — x)(s — y)(s — z),
where s = j(x + y + z) is the so-called semi-
perimeter of the triangle. Use Heron's formula to show
that, for a fixed perimeter P, the triangle with the
largest area is equilateral.
29. Use a Lagrange multiplier to find the largest sphere
centered at the origin that can be inscribed in the ellip-
soid 3x2 + 2y2 + z2 = 6. (Be careful with this prob-
lem; drawing a picture may help.)
30. Find the point closest to the origin and on the line
of intersection of the planes 2x + y + 3z = 9 and
3x + 2y + z = 6.
292 Chapter 4 I Maxima and Minima in Several Variables
31 . Find the point closest to the point (2,5,-1) and on the
line of intersection of the planes x — 2y + 3z = 8 and
2z-y = 3.
32. The plane x + y + z = 4 intersects the paraboloid
z = x2 + y2 in an ellipse. Find the points on the el-
lipse nearest to and farthest from the origin.
33. Find the highest and lowest points on the ellipse ob-
tained by intersecting the paraboloid z
the plane x + y + 1z = 2,
+ yz with
34. Find the minimum distance between a point on the
ellipse x2 + 2y2 = 1 and a point on the line x + y = 4.
(Hint: Consider a point (x , y ) on the ellipse and a point
(u, v) on the line. Minimize the square of the distance
between them as a function of four variables. This prob-
lem is difficult to solve without a computer.)
35. (a) Use the method of Lagrange multipliers to find crit-
ical points of the function f(x, y) = x + y subject
to the constraint xy = 6.
(b) Explain geometrically why / has no extrema on
the set {(x, y) \ xy = 6}.
36. Leta, fi, and y denote the (interior) angles ofatriangle.
Determine the maximum value of sin a sin /3 sin y .
37. Let S be a surface in R3 given by the equation
g(x, y, z) = c, where g is a function of class C1 with
nonvanishing gradient and c is a constant. Suppose that
there is a point P on S whose distance from the origin
is a maximum. Show that the displacement vector from
the origin to P must be perpendicular to S.
38. The cylinder x2 + y2 = 4 and the plane 2x + 2y +
z = 2 intersect in an ellipse. Find the points on
the ellipse that are nearest to and farthest from the
origin.
39. Find the points on the ellipse 3x2 — Axy + 3y2 = 50
that are nearest to and farthest from the origin.
40. This problem concerns the determination of the ex-
trema of f(x, y) = ~Jx + subject to the con-
straint x2 + y2 = 17, where x > 0 and y > 0.
(a) Explain why / must attain both a global minimum
and a global maximum on the given constraint
curve.
(b) Use a Lagrange multiplier to solve the system of
equations
\Vf(x,y) = XVg{x,y)
U(x, y) = o
where g(x, y) = x2 + y2. You should identify a
single critical point of /.
(c) Identify the global minimum and the global max-
imum of / subject to the constraint.
41 . Consider the problem of finding extrema of f(x, y) =
x subject to the constraint y2 — 4x3 + 4x4 = 0.
(a) Use a Lagrange multiplier and solve the system of
equations
Wf(x,y) = Wg(x,y)
[g(x,y) = 0
where g(x, y) = y2 — 4x3 + 4xA . By doing so,
you will identify critical points of / subject to the
given constraint.
(b) Graph the curve y2 — 4x3 + 4x4 = 0 and use the
graph to determine where the extrema of f(x, y) =
x occur.
(c) Compare your result in part (a) with what you
found in part (b). What accounts for any differ-
ences that you observed?
42. Consider the problem of finding extrema of
f(x, y, z) = x2 + y2 subject to the constraint z = c,
where c is any constant.
(a) Use the method of Lagrange multipliers to iden-
tify the critical points of / subject to the constraint
given above.
(b) Using the usual alphabetical ordering of variables
(i.e., x\ =x,x2 = y, x^ = z), construct the Hessian
matrix HL(X;a\, 02, 03) (where L(l;x, y, z) =
f{x, y, z) — l(z — c)) for each critical point you
found in part (a). Try to use the second deriva-
tive test for constrained extrema to determine the
nature of the critical points you found in part (a).
What happens?
(c) Repeat part (b), this time using the variable order-
ing x\ = z, X2 = y, X3 = x. What does the second
derivative test tell you now?
(d) Without making any detailed calculations, discuss
why / must attain its minimum value at the point
(0, 0, c). Then try to reconcile your results in parts
(b) and (c). This exercise demonstrates that the as-
sumption that
det
dx
dx
(a)
(a)
dxk
dxk
(a)
(a)
#0
is important.
43. Consider the problem of finding critical points of the
function f(xi, ■ ■ ■ , x„) subject to the set of k con-
straints
gl(Xl, ...,x„) = ci, gi{xu ...,*„) = c2, ... ,
gk(x\, ...,x„) = ck.
Assume that f, gi, g2, • ■ ■ , gk are all of class C2.
4.4 | Some Applications of Extrema
(a) Show that we can relate the method of Lagrange
multipliers for determining constrained critical
points to the techniques in §4.2 for finding un-
constrained critical points as follows: If
(k, a) = (ki, . . . , kk;au . . . , an)
is a pair consisting of k values for Lagrange mul-
tipliers k\, . . . , kk and n values a\ , . . . , a„ for the
variables X\, .. . ,x„ such that a is a constrained
critical point, then (k, a) is an ordinary (i.e., un-
constrained) critical point of the function
LQi, ...,lk;xi, ...,x„)
k
= f(X\, ...,Xn)~ } li(gi(Xl, ... ,Xn)- Ci).
1=1
(b) Calculate the Hessian HL(k, a), and verify that it
is the matrix used in §4.3 to provide the criterion
for determining the nature of constrained critical
points.
44. The unit hypersphere in R" (centered at the origin
0 = (0, . . . , 0)) is defined by the equation xj+xj +
■ ■ ■ + x\ = 1. Find the pair of points x = (xi xn)
and y = (vi, . • • , y„), each of which lies on the unit
hypersphere, that maximizes and minimizes the func-
tion
n
f(x\, Xn, Vi, . . • , y„) = xtyt.
i=l
What are the maximum and minimum values of /?
45. Let x = (xi, . . . , Xn) and y = (y\, . . . , yn) be any vec-
tors in R" and, for = 1, . . . , n, set
u; = = and v, = =.
(a) Show that u = («i , . . . , un) and v = (i>i , . . . , vn)
lie on the unit hypersphere in R" .
(b) Use the result of Exercise 44 to establish the
Cauchy-Schwarz inequality
|x-y| < II x || ||y||.
4.4 Some Applications of Extrema
In this section, we present several applications of the methods for finding both
constrained and unconstrained extrema discussed previously.
53
X
Protein level
Figure 4.33 Height versus protein
level.
Protein level
Figure 4.34 Fitting a line to
the data.
Least Squares Approximation
The simplest relation between two quantities x and y is, without doubt, a linear
one: y = mx + b (where m and b are constants). When a biologist, chemist,
psychologist, or economist postulates the most direct connection between two
types of observed data, that connection is assumed to be linear. Suppose that Bob
Biologist and Carol Chemist have measured certain blood protein levels in an
adult population and have graphed these levels versus the heights of the subjects
as in Figure 4.33. If Prof. Biologist and Dr. Chemist assume a linear relationship
between the protein and height, then they desire to pass a line through the data as
closely as possible, as suggested by Figure 4.34.
To make this standard empirical method of linear regression precise (in-
stead of merely graphical and intuitive), we first need some notation. Suppose we
have collected n pairs of data (x\ , yi), (x2, yi), • ■ ■ , (xn , y„ ). (In the example just
described, xt is the protein level of the ith subject and y, his or her height.) We
assume that there is some underlying relationship of the form y = mx + b, and
we want to find the constants m and b so that the line fits the data as accurately
as possible. Normally, we use the method of least squares. The idea is to find
the values of m and b that minimize the sum of the squares of the differences
between the observed y-values and those predicted by the linear formula. That is,
we minimize the quantity
D(m, b) = [vi - {mx\ + b)f + [y2 - (mx2 + b)f
+ --- + [yn-(mxn+b)f,
(1)
294 Chapter 4 | Maxima and Minima in Several Variables
Line y = mx + b
Figure 4.35 The method of least squares.
where, for i = 1, . . . , n, y,- represents the observed y-value of the data, and
mxi + b represents the y-value predicted by the linear relationship. Hence, each
expression in D of the form y, — (mx, + b) represents the error between the
observed and predicted y-values. (See Figure 4.35.) They are squared in the
expression for D in order to avoid the possibility of having large negative
and positive terms cancel one another, thereby leaving little or no "net error,"
which would be misleading. Moreover, D(m , b) is the square of the distance
in R" between the point (yi, y2, . . . , y„) and the point {mx\ + b, mx2 + b, . . . ,
mxn + b).
Thus, we have an ordinary minimization problem at hand. To solve it, we
need to find the critical points of D. First, we can rewrite D as
n
D(m, b) = y^[yj - (mxi + b)]2
n n n n
= ^2 yf - 2m X! Xtyi ~2b^2, yi + XI (mxi + fo)2-
i=i i=i i=i
H ii
1=1 1=1
ii n n
2 ^2 xi Vi + 2m xf + 2b x\
i=l i=l 1 = 1
ii n
-- -2 yt + X! 2{,nXi + b">
1 = 1 i=l
11 17
= —2 y,- + 2m X; + 2nb.
i = l i=l
When we set both partial derivatives equal to zero, we obtain the following pair
of equations, which have been simplified slightly:
{H x?) m + (H xi)b = H x>yi
(2)
(J2xj)m + nb = J2yi
(All sums are taken from i = 1 to «.) Although (2) may look complicated, it is
nothing more than a linear system of two equations in the two unknowns m and b.
Then
i=i
dD
dm
and
3D
~db
4.4 | Some Applications of Extrema
It is not difficult to see that system (2) has a single solution. Therefore, we have
shown the following:
PROPOSITION 4.1 Given n data points (jci, yi), (x2, yi), . . . , (x„, y„) with not
all of x\, X2, . . ■ , xn equal, the function
n
D(m,b) = J2[yl-(mxl+b)f
i=i
has a single critical point (mo, bo) given by
mo = j '
and
C>0 = 2 "
Since D(m , b) is a quadratic polynomial in m and b, the graph of z = D(m,b)
is a quadric surface. (See §2. 1 .) The only such surfaces that are graphs of functions
are paraboloids and hyperbolic paraboloids. We show that, in the present case,
the graph is that of a paraboloid by demonstrating that D has a local minimum at
the critical point (mo, &o) given in Proposition 4.1.
We can use the Hessian criterion to check that D has a local minimum at
(mo, bo). We have
HD(m, b) =
2 ^2 Xj 2n
The principal minors are and An J2xf ~ 4(X>,) • The first minor is
obviously positive, but determining the sign of the second requires a bit more
algebra. (If you wish, you can omit reading the details of this next calculation
and rest assured that the story has a happy ending.) Ignoring the factor of 4, we
examine the expression n xf — (J2 *;)2- Expanding the second term yields
»£*?-(£■
= n J2x> - [ Ex'2 + J22x'xj )
i=i \ i=i i<j )
n
= {n - l)^x'2 ~ ^22xixJ-
i'=l i<j
On the other hand we have
n
y^xxi ~ xj^2 = £ (xf ~ 2x'xj + x)) = (« - 1) £ xf - £ 2xjXj.
(To see that equation (4) holds, you need to convince yourself that
£tf + *5) = 0i-i)I>?
(3)
(4)
296 Chapter 4 | Maxima and Minima in Several Variables
by counting the number of times a particular term of the form x\ appears in the
left-hand sum.) Thus, we have
det HD(m,b) = 4 I n^x2 - f^*,
Best fit
line
* (5, 4)
Figure 4.36 Data for the linear
regression of Example 1.
= 4 1 (n — 1) x2 — IxiXj j by equation (3),
\ i=i <</' /
= 4 / ,(x,- — Xj)2 by equation (4).
KJ
Because this last expression is a sum of squares, it is nonnegative. Therefore, the
Hessian criterion shows that D does indeed have a local minimum at the critical
point. Hence, the graph of z = D(m, b) is that of a paraboloid. Since the (unique)
local minimum of a paraboloid is in fact a global minimum (consider a typical
graph), we see that D is indeed minimized at (m0, b0).
EXAMPLE 1 To see how the preceding discussion applies to a specific set of
data, consider the situation depicted in Figure 4.36.
We have n = 5, and the function D to be minimized is
D(m, b) = [2- (m + b)f + [1 - (2m + b)f + [5 - (3m + b)]2
+ [3 - (4m + b)f + [4 - (5m + b)f.
We compute
= 15,
Thus, using Proposition 4.1,
5-51-15-15
m
5-55-15-15
55,
_ 3
" 5'
15,
55 ■ 15 - 15 ■ 51 _ 6
5-55-15-15 ~ 5'
The best fit line in terms of least squares approximation is
3 6
V = -x H — .
"^5 5
Of course, linear regression is not always an appropriate technique. It may
not be reasonable to assume that the data points fall nearly on a straight line.
Some formula other than y = mx + b may have to be assumed to describe the
data with any accuracy. Such a postulated relation might be quadratic,
y = ax2 + bx + c,
or x and y might be inversely related
a
y=-+b.
x
You can still apply the method of least squares to construct a function analogous
to D in equation (1) to find the relation of a given form that best fits the data.
Another way that least squares arise is if y depends not on one variable but
on several: x\, X2, ■ ■ ■ , xn. For example, perhaps adult height is measured against
4.4 | Some Applications of Extrema
Figure 4.37 A particle traveling
in a force field F.
blood levels of 10 different proteins instead of just one. Multiple regression is
the statistical method of finding the linear function
y = a\X\ + a2x2 +
that best fits a data set of (n + l)-tuples
+ a„xn + b
(a-
(i)
x1
,xil\ yi),(*r.*2
(2) (2)
{x
(*) VW
1 ' x2 '
*,fU)
We can find such a "best fit hyperplane" by minimizing the sum of the squares of
the differences between the y-values furnished by the data set and those predicted
by the linear formula. We leave the details to you.3
Physical Equilibria
Let F: X c R3 — > R3 be a continuous force field acting on a particle that moves
along a path x: / c R —> R3 as in Figure 4.37. Newton's second law of motion
states that
F(x(f)) = mx"(t),
(5)
where m is the mass of the particle. For the remainder of this discussion, we will
assume that F is a gradient field, that is, that F = — VV for some C1 potential
function V:X c R3 -> R. (See §3.3 for a brief comment about the negative sign.)
We first establish the law of conservation of energy.
THEOREM 4.2 (Conservation of energy) Given the set-up above, the
quantity
\m\\x!{t)\\2 + V(x(0)
is constant.
The term hn ||x'(?) ||2 is usually referred to as the kinetic energy of the particle
and the term V(x(r)) as the potential energy. The significance of Theorem 4.2
is that it states that the sum of the kinetic and potential energies of a particle
is always fixed (conserved) when the particle travels along a path in a gradient
vector field. For this reason, gradient vector fields are also called conservative
vector fields.
Proof of Theorem 4.2 As usual, we show that the total energy is constant by
showing that its derivative is zero. Thus, using the product rule and the chain rule,
we calculate
d r,
dt L
i'(0 • x'(0 + V(x(/))] = mx"(t) ■ x'(t) + V V(x(0) • x'(0
= mx"(t) • x'(t) - F(x(0) • x'(t)
= mx"(t) • x'(t) - mx"(t) • x'(t)
= 0,
from the definitions of F and V and by formula (5).
3 Or you might consult S.Weisberg, Applied Linear Regression, 2nd ed., Wiley-Interscience, 1985,
Chapter 2. Be forewarned, however, that to treat multiple regression with any elegance requires somewhat
more linear algebra than we have presented.
298 Chapter 4 | Maxima and Minima in Several Variables
Figure 4.38 For a stable
equilibrium point, the path of a
nearby particle with a sufficiently
small kinetic energy will remain
nearby with a bounded kinetic
energy.
In physical applications it is important to identify those points in space that
are "rest positions" for particles moving under the influence of a force field. These
positions, known as equilibrium points, are such that the force field does not act
on the particle so as to move it from that position. Equilibrium points are of two
kinds: stable equilibria, namely, equilibrium points such that a particle perturbed
slightly from these positions tends to remain nearby (for example, a pendulum
hanging down at rest) and unstable equilibria, such as the act of balancing a ball
on your nose. The precise definition is somewhat technical.
DEFINITION 4.3 Let F: X c R" -> R" be any force field. Then x0 e X is
called an equilibrium point of F if F(xo) = 0. An equilibrium point xo is
said to be stable if, for every r, e > 0, we can find other numbers ro, eo > 0
such that if we place a particle at position x with || x — x0 1| < r0 and provide it
with a kinetic energy less than eo, then the particle will always remain within
distance r of xq with kinetic energy less than e.
In other words, a stable equilibrium point xo has the following property: You
can keep a particle inside a specific ball centered at xo with a small kinetic energy
by starting the particle inside some other (possibly smaller) ball about x0 and
imparting to it some (possibly smaller) initial kinetic energy. (See Figure 4.38.)
THEOREM 4.4 For a C1 potential function V of a vector field F = -V V,
1. The critical points of the potential function are precisely the equilibrium
points of F.
2. If x0 gives a strict local minimum of V, then x0 is a stable equilibrium point
ofF.
EXAMPLE 2 The vector field F = (-6x - 2y - 2)i + {-2x - Ay + 2)j is
conservative and has
V(x, y) = 3x2 + 2xy + 2x + 2y2 - 2y + 4
as a potential function (meaning that F = — V V, according to our current sign
convention). There is only one equilibrium point, namely, (— | , |) . To see if it is
stable, we look at the Hessian of V :
6 2
2 4
The sequence of principal minors is 6, 20. By the Hessian criterion, (— |, |) is a
strict local minimum of V and, by Theorem 4.4, it must be a stable equilibrium
point of F. ♦
Proof of Theorem 4.4 The proof of part 1 is straightforward. Since F = — VV,
we see that F(x) = 0 if and only if W(x) = 0. Thus, equilibrium points of F are
the critical points of V.
To prove part 2, let xo be a strict local minimum of V and x:/^R"aC'
path such that x(?0) = x0 for some to e /. By conservation of energy, we must
have, for all t 6 /, that
im||x'(0H2 + V(x(/)) = jm\\x'(t0)\\2 + V(x(t0)).
4.4 | Some Applications of Extrema
Figure 4.39 On the surface
S = {x ] g{\) = c], the component
of F that is tangent to S at x is
denoted by $(x).
To show that xo is a stable equilibrium point, we desire to show that we can bound
the distance between x(/) and xo = x(fo) by any amount r and the kinetic energy
by any amount e. That is, we want to show we can achieve
||x(O-x0|| < r
(i.e., x(?) e Br(x0) in the notation of §2.2) and
im||x'(OH2 < e.
As the particle moves along x away from xo, the potential energy must increase
(since xo is assumed to be a strict local minimum of potential energy), so the
kinetic energy must decrease by the same amount. For the particle to escape from
B, (x0), the potential energy must increase by a certain amount. If eo is chosen to
be smaller than that amount, then the kinetic energy cannot decrease sufficiently
(so that the conservation equation holds) without becoming negative. This being
clearly impossible, the particle cannot escape from Br(xo). ■
Often a particle is not only acted on by a force field but also constrained to lie
in a surface in space. The set-up is as follows: F is a continuous vector field on R3
acting on a particle that lies in the surface S = {x 6 R3 | g(x) = c}, where g is a
C1 function such that Vg(x) ^ 0 for all x in S. Most of the comments made in the
unconstrained case still hold true, provided F is replaced by the vector component
of F tangent to S. Since, at x e S, Vg(x) is normal to S, this tangential component
of F at x is
<Kx) = F(x) - projVi?(x)F(x). (6)
(SeeFigure4.39.)Theninplaceofformula(5),wehave,forapathx: / cR->S,
$(x(f)) = mx"(t). (7)
We can now state a "constrained version" of Theorem 4.4.
THEOREM 4.5 For a C1 potential function V of a vector field F = - V V,
1. If V|s has an extremum at xo e 5, then xo is an equilibrium point in S.
2. If V | s has a strict local minimum at x0 e S, then x0 is a stable equilibrium
point.
Sketch of proof Forpart 1, if V\s has an extremum at xo, then, by Theorem 3.1,
we have, for some scalar X, that
VV(x0) = AVg(x0).
Hence, because F = — VV,
F(x0) = -AVg(xo),
implying that F is normal to 5 at xo. Thus, there can be no component of F tangent
to S at xo (i.e., 4>(xo) = 0). Since the particle is constrained to lie in S, we see
that the particle is in equilibrium in S.
The proof of part 2 is essentially the same as the proof of part 2 of Theorem
4.4. The main modification is that the conservation of energy formula in Theorem
4.2 must be established anew, as its derivation rests on formula (5), which has
been replaced by formula (7). Consequently, using the product and chain rules,
Chapter 4 i Maxima and Minima in Several Variables
x2 + y2 + (z - 2r)2
(0, 0, 3r)
= r2-i
f F '
'(0,0./-)
Figure 4.40 On the sphere
x2 + y2 + (z- 2r)2 = r2, the
points (0, 0, r) and (0, 0, 3r)
are equilibrium points for the
gravitational force field
F = — mgk.
we check, for x: /
d
dt
m\\x'(t)\\2 + V(x(r))] = jt[^mx'(t).x'(t)+ V(x(t))]
= mx"(t) • x'(t) + VV(x(/)) • x'(0-
Then, using formula (6), we have
d
dt
m \\x'(t)\\2 + V(x(0)] = x'(t) • mx"(t) - F(x(f )) • x'(t)
= x'(t).<t>(x(t))-F(x(t)).x'(t)
= x'(t) ■ [F(x(0) - projVg(x(r))F(x(0)]
-F(x(/))-x'(f)
= -x'(0-projVs(x(r))F(x(0)
after cancellation. Thus, we conclude that
d
dt
since x'(t ) is tangent to the path in S and, hence, tangent to S itself at x(t), while
projvo(X(j))F(x(0) is parallel to Vg(x(f )) and, hence, perpendicular to S at x(r). ■
m||x'(OII2 + V(x(0)] = 0,
EXAMPLE 3 Near the surface of the earth, the gravitational field is ap-
proximately
F = —mgk.
(We're assuming that, locally, the surface of the earth is represented by the plane
z = 0.) Note that F = -V V, where
V(x, y, z) = mgz.
Now suppose a particle of mass m lies on a small sphere with equation
h(x, y, z) = x2 + y2 + (z- 2rf = r2.
We can find constrained equilibria for this situation, using a Lagrange multiplier.
The gradient equation V V = XV h, along with the constraint, yields the system
0 = 2Ax
0 = 2ky
mg = 2X(z — 2r)
x2 + y2 + (z - 2r)2 = r2
Because m and g are nonzero, X cannot be zero. The first two equations imply
x = y = 0. Therefore, the last equation becomes
which implies
(z-2r)2 = r2,
Z = r, 3r
are the solutions. Consequently, the positions of equilibrium are (0, 0, r) and
(0, 0, 3r) (corresponding to X = —mg/2r and +mg/2r, respectively). From ge-
ometric considerations, we see V is strictly minimized at S at (0, 0, r) and maxi-
mized at (0, 0, 3r) as shown in Figure 4.40. From physical considerations, (0, 0, r)
4.4 | Some Applications of Extrema
is a stable equilibrium and (0, 0, 3r) is an unstable one. (Try balancing a marble
Applications to Economics
We present two illustrations of how Lagrange multipliers occur in problems in-
volving economic models.
EXAMPLE 4 The usefulness of amounts x\, x%, . . . , xn of (respectively) dif-
ferent capital goods G\,Gi, ...,Gn can sometimes be measured by a function
U(x\, X2, ■ . . , xn), called the utility of these goods. Perhaps the goods are indi-
vidual electronic components needed in the manufacture of a stereo or computer,
or perhaps U measures an individual consumer's utility for different commodi-
ties available at different prices. If item G, costs a, per unit and if M is the total
amount of money allocated for the purchase of these n goods, then the consumer
or the company needs to maximize U(x\, x% x„) subject to
This is a standard constrained optimization problem that can readily be approached
by using the method of Lagrange multipliers.
For instance, suppose you have a job ordering stationery supplies for an office.
The office needs three different types of products a, b, and c, which you will order
in amounts x, y, and z, respectively. The usefulness of these products to the smooth
operation of the office turns out to be modeled fairly well by the utility function
U(x, y, z) = xy + xyz. If product a costs $3 per unit, product b $2 a unit, and
product c $ 1 a unit and the budget allows a total expenditure of not more than
$899, what should you do? The answer should be clear: You need to maximize
U(x, y, z) = xy + xyz subject to B(x, y, z) = 3x + 2y + z = 899.
The Lagrange multiplier equation, VC/(x, y, z) = XVB(x, y, z), and the budget
constraint yield the system
The last equality implies that either x = 0 or y = (z + l)/2. We can reject the
first possibility, since t/(0, y, z) = 0 and the utility U(x, y, z) > 0 whenever
x, y, and z are all positive. Thus, we are left with y = (z + l)/2. This in turn
implies that A. = (z + l)2/6. Substituting for y in the constraint equation shows
that x = (898 — 2z)/3, so that equation xy = X becomes
on top of a ball.)
♦
a^xi + 02X2 + ■ ■ ■ + a„xn = M.
y + yz = 3A
x + xz = 2k
xy = X
3x + 2y + z = 899
Solving for X in the first three equations yields
which is satisfied by either z = — 1 (which we reject) or by z = 299. The only
realistic critical point for this problem is (100, 150, 299). We leave it to you to
check that this point is indeed the site of a maximum value for the utility. ♦
Chapter 4 i Maxima and Minima in Several Variables
EXAMPLE 5 In 1928, C.W.Cobb and P. M.Douglas developed a simple model
for the gross output Q of a company or a nation, indicated by the function
Q(K,L) = AK"L1-",
where K represents the capital investment (in the form of machinery or other
equipment), L the amount of labor used, and A and a positive constants with
0 < a < 1 . (The function Q is known now as the Cobb-Douglas production
function.) If you are president of a company or nation, you naturally wish to
maximize output, but equipment and labor cost money and you have a total
amount of M dollars to invest. If the price of capital is p dollars per unit and
the cost of labor (in the form of wages) is w dollars per unit, so that you are
constrained by
B(K, L) = pK + wL < M,
what do you do?
Again, we have a situation ripe for the use of Lagrange multipliers. Before
we consider the technical formalities, however, we consider a graphical solution.
Draw the level curves of Q, called isoquants, as in Figure 4.41. Note that Q
increases as we move away from the origin in the first quadrant. The budget
constraint means that you can only consider values of K and L that lie inside or
on the shaded triangle. It is clear that the optimum solution occurs at the point
(K, L) where the level curve is tangent to the constraint line pK + wL = M.
Here is the analytical solution: From the equation VQ(K, L) = XVB(K, L)
plus the constraint, we obtain the system
'AaK"-lLl-a = Xp
A(l - d)Kalra = Xw.
pK + wL = M
Solving for p and w in the first two equations yields
K
Figure 4.41 A family of isoquants. The optimum value of Q(K, L) subject to the
constraint pK + wL = M occurs where a curve of the form Q = c is tangent to the
constraint line.
4.4 I Exercises 303
Substitution of these values into the third equation gives
Aa „ i „ A(l - a) „ , „
— K L + — -K"Ll-a = M.
X X
Thus,
X = —KaL1-",
M
and the only critical point is
(Ma M(\-a)
\ P w
From this geometric discussion, we know that the critical point must yield the
maximum output Q.
From the Lagrange multiplier equation, at the optimum values for L and K,
we have
p dK w dL
This relation says that, at the optimum values, the marginal change in output per
dollar's worth of extra capital equals the marginal change per dollar's worth of
extra labor. In other words, at the optimum values, exchanging labor for capital
(or vice versa) won't change the output. This is by no means the case away from
the optimum values.
There is not much that is special about the function Q chosen. Most of our
observations remain true for any C2 function Q that satisfies the conditions
dQ dQ >Q d2Q d2Q
dK' dL ~ ' dK2, dL2 <
If you consider what these relations mean qualitatively about the behavior of the
output function with respect to increases in capital and labor, you will see that
they are entirely reasonable assumptions.4 ♦
4.4 Exercises
1. Find the line that best fits the following data: (0, 2),
(1,3), (2,5), (3,3), (4,2), (5,7), (6,7).
2. Show that if you have only two data points (xi , y\ ) and
(xi, yi), then the best fit line given by the method of
least squares is, in fact, the line through (x\, yi) and
(xz, yi)-
3. Suppose that you are given « pairs of data (xi, y\),
(X2, yz), ■ ■ ■ , (x„, yn) and you seek to fit a function of
the form y = a/x + b to these data.
(a) Use the method of least squares as outlined in this
section to construct a function D(a, b) that gives
the sum of the squares of the distances between
observed and predicted j-values of the data.
(b) Show that the "best fit" curve of the form y =
a/x + b should have
n^yi/xt -(E !/*)(£*■)
2
»E i/*?-(£i/*)
and
b _ (E i/-*,2) (E yj) - (E im) (E y.M-)
»EV*?-(Ei/*)2
(All sums are from i = 1 to «.)
4 For more about the history and derivation of the Cobb— Douglas function, consult R. Geitz, "The Cobb-
Douglas production function," UMAP Module No. 509, Birkhauser, 1981.
304 Chapter 4 | Maxima and Minima in Several Variables
4. Find the curve of the form y = a/x + b that best fits
the following data: (1 , 0), (2, - 1), (|, 1), and (3, - 1) .
(See Exercise 3.)
5. Suppose that you have n pairs of data (jci , ji),
(xi, yi), ■ ■ ■ , (xn, yn) and you desire to fit a quadratic
function of the form y = ax2 + bx + c to the data.
Show that the "best fit" parabola must have coefficients
a, b, and c satisfying
(E xty + (E*,3)& + (E-*,2)c = Z*fyi
(E -v,3)« + (E 4)b + (E *)c = E xm .
(E^2)« + (E^)^ + «c = E.v/
(All sums are from i = 1 to n.)
6. (Note: This exercise will be facilitated by the use of
a spreadsheet or computer algebra system.) Egbert
recorded the number of hours he slept the night before
a major exam versus the score he earned, as shown in
the table below.
(a) Find the line that best fits these data.
(b) Find the parabola y = ax2 + bx + c that best fits
these data. (See Exercise 5.)
(c) Last night Egbert slept 6.8 hr. What do your an-
swers in parts (a) and (b) predict for his score on
the calculus final he takes today?
Hours of sleep
Test score
8
85
8.5
72
9
95
7
68
4
52
8.5
75
7.5
90
6
65
7. Let F = {-2x - 2y - l)i + (-2x - 6y - 2)j.
(a) Show that F is conservative and has potential
function
V(x, y) = x2 + 2xy + 3y2 +x + 2y
(i.e.,F = -VV).
(b) What are the equilibrium points of F? The stable
equilibria?
8. Suppose a particle moves in a vector field F in R2 with
physical potential
V(x, y) = 2x2 - Sxy - y2 + I2x - 8y + 12.
Find all equilibrium points of F and indicate which, if
any, are stable equilibria.
9. Let a particle move in the vector field F in R3 whose
physical potential is given by
V(x, y, z) = 3x2 + 2xy + z2 - 2yz + 3x + 5y - 10.
Determine the equilibria of F and identify those that
are stable.
10. Suppose that a particle of mass m is constrained to
move on the ellipsoid 2x2 + 3y2 + z2 = 1 subject to
both a gravitational force F = —mgk., as well as to an
additional potential V(x, y, z) = 2x.
(a) Find any equilibrium points for this situation.
(b) Are there any stable equilibria?
1 1 . The Sukolux Vacuum Cleaner Company manufactures
and sells three types of vacuum cleaners: the standard,
executive, and deluxe models. The annual revenue in
dollars as a function of the numbers x, y, and z (re-
spectively) of standard, executive, and deluxe models
sold is
R{x, y, z) = xyz2 - 25,000* - 25,000y - 25,000z.
The manufacturing plant can produce 200,000 total
units annually. Assuming that everything that is manu-
factured is sold, how should production be distributed
among the models so as to maximize the annual
revenue?
1 2. Some simple electronic devices are to be designed to
include three digital component modules, types 1, 2,
and 3, which are to be kept in inventory in respective
amounts X\, x%, and xj,. Suppose that the relative im-
portance of these components to the various devices is
modeled by the utility function
U (x\ , X%, X3) = X\X2 + 2x\X^ + X\X2Xt,.
You are authorized to purchase $90 worth of these parts
to make prototype devices. If type 1 costs $1 per com-
ponent, type 2 $4 per component, and type 3 $2 per
component, how should you place your order?
13. A farmer has determined that her cornfield will yield
corn (in bushels) according to the formula
B(x,y) = 4x2 + y2 + 600,
where x denotes the amount of water (measured in
hundreds of gallons) used to irrigate the field and y
the number of pounds of fertilizer applied to the field.
The fertilizer costs $10 per pound and water costs $15
per hundred gallons. If she can allot $500 to prepare
her field through irrigation and fertilization, use a
Lagrange multiplier to determine how much water and
fertilizer she should purchase in order to maximize her
yield.
14. A textile manufacturer plans to produce a cashmere/
cotton fabric blend for use in making sweaters. The
amount of fabric that can be produced is given by
f(x,y) = 4xy-2x-8y + 3,
True/False Exercises for Chapter 4
where x denotes the number of pounds of raw cash-
mere used is and y is the number of pounds of raw
cotton. Cotton costs $2 per pound and cashmere costs
$8 per pound.
(a) If the manufacturer can spend $ 1 000 on raw mate-
rials, use a Lagrange multiplier to advise him how
he should adjust the ratio of materials in order to
produce the most cloth.
(b) Now suppose that the manufacturer has a budget
of B dollars. What should the ratio of cotton to
cashmere be (in terms of 5)? What is the limiting
value of this ratio as B increases?
15. The CEO of the Wild Widget Company has decided
to invest $360,000 in his Michigan factory. His eco-
nomic analysts have noted that the output of this
factory is modeled by the function Q(K, L) =
60 K1^ L2/3 , where K represents the amount (in thou-
sands of dollars) spent on capital equipment and L
represents the amount (also in thousands of dollars)
spent on labor.
(a) How should the CEO allocate the $360,000
between labor and equipment?
(b) Check that BQ/dK = dQ/dL at the optimal
values for K and L.
16. Let Q{K,L) be a production function for a com-
pany where K and L represent the respective amounts
spent on capital equipment and labor. Let p denote the
price of capital equipment per unit and w the cost of
labor per unit. Show that, subject to a fixed produc-
tion Q(K, L) = c, the total cost M of production is
minimized when K and L are such that
j_3G _ j_ae
p dK w dL
True/False Exercises for Chapter 4
1 . If / is a function of class C2 and p% denotes the second-
order Taylor polynomial of / at a, then /(x) ~ P2W
when x ~ a.
2. The increment A/ of a function f(x, y) measures the
change in the z-coordinate of the tangent plane to the
graph of /.
3. The differential df of a function f(x, y) measures the
change in the z-coordinate of the tangent plane to the
graph of /.
4. The second-order Taylor polynomial of f(x, y, z) =
x2 + 3xz + y2 at (1, —1, 2) is pi(x, y, z) = x2 + 3xz
+ y2-
5. The second-order Taylor polynomial of f(x,y) =
x3 + 2xy + y at (0, 0) is pi(x, y) = 2xy + y.
6. The second-order Taylor polynomial of f(x,y) =
x3 + 2xy + y at (1, —1) is p2(x, y) = 2xy + y.
7. Near the point (1,3,5), the function f(x,y,z) =
3x4 + 2y3 + z2 is most sensitive to changes in z.
8. The Hessian matrix Hf(x\ , . . . , xn) of / has the prop-
erty that Hf(Xl ,...,xn)T = Hf(xx x„),
9. If V/(ai , . . . , an) = 0, then / has a local extremum
at a = (ai, . . . , a„).
10. If / is differentiable and has a local extremum at
a = (ai, . . . , an), then V/(a) = 0.
11. The set {(x, y, z) \ 4 < x2 + y2 + z2 < 9} is compact.
12. The set {(x, y) \ 2x — 3y = 1} is compact.
13. Any continuous function f(x, y) must attain a global
maximum on the disk {(x, y) \ x2 + y2 < 1}.
14. Any continuous function f(x,y,z) must attain a
global maximum on the ball {(x, y, z) | (x — l)2 +
(y+l)2 + z2<4).
15. If f(x, y) is of class C2, has a critical point at (a, b),
and fxx(a, b)fyy(a, b) — fxy(a, b)2 < 0, then / has a
saddle point at (a, b).
16. If det ///(a) = 0, then / has a saddle point at a.
17. The function f(x, y, z) = xiy2z — x2(y + z) has a
saddle point at (1 , — 1 , 2).
18. The function f(x, y, z) = x2 + y2 + z2 — yz has a lo-
cal maximum at (0, 0, 0).
19. The function f(x, y, z) = xy3 — x2z + z has a degen-
erate critical point at (— 1 , 0, 0).
20. The function F(xx ,...,.«„) = 2{xx - l)2 - 3(x2 - 2)2
_l_ 1_ (— iyi+1(n + l)(xn — rif has a critical point at
(l,2,...,n).
21 . The function F(xi, x„) = 2(xi- l)2 - 3(x2 - 2)2
+ ■ ■ ■ + (— l)"+1(rc + l)(xn — n)2 has a minimum at
(l,2,...,n).
22. All local extrema of a function of more than one vari-
able occur where all partial derivatives simultaneously
vanish.
23. All points a = («i, . . . , ai) where the function
f(x\, ... ,x„) has an extremum subject to the con-
straint that g{x\, ... ,xn) — c, are solutions to the
Chapter 4 ! Maxima and Minima in Several Variables
system of equations
9/
3.vi
9xi
X -
24. Any solution (Ai
equations
9*1
9x„ 9x„
g(Xi, ...,x„) = c
. . . , Xk,x\, . . . , x„) to the system of
dx\ dx\
= Ai + • • • + Ajt
dxn dx„ dx„
gi(xu ■■, X„) = C\
gl(xi, ...,Xn) = Ck
yields a point (x\, . . . , xn) that is an extreme value
of / subject to the simultaneous constraints g\ =
a, ...,gk = ck.
25. To find the critical points of the function f(x, y, z, w)
subject to the simultaneous constraints g{x, y, z, w) =
c, h(x, y , z, w) = d, k(x, y,z,w) = e using the tech-
nique of Lagrange multipliers, one will have to solve
a system of four equations in four unknowns.
26. Suppose that f(x, y, z) and g(x, y, z) are of class C1
and that (xo , yo , zo) is a point where / achieves a maxi-
mum value subject to the constraint that g(x , y , z) = c
and that Vg(xo, yo, Zo) is nonzero. Then the level set
of / that contains (xo, yo, zo) must be tangent to the
level set S = {(x, y, z) \ g{x, y, z) = c}.
27. The critical points of f(x, y, z) = xy + 2xz + 2yz
subject to the constraint that xyz = 4 are the same
as the critical points of the function F(x, y) =
8 8
xv + - + -.
x y
28. Given data points (3, 1), (4, 10), (5, 8), (6, 12), to find
the best fit line by regression, we find the minimum
value of the function D(m, b) = (3m + b — l)2 +
(4m + b - 10)2 + (5m + b - 8)2 + (6m + b - 12)2.
29. All equilibrium points of a gradient vector field
are minimum points of the vector field's potential
function.
30. Given an output function for a company, the marginal
change in output per dollar investment in capital is the
same as the marginal change in the output per dollar
investment in labor.
Miscellaneous Exercises for Chapter 4
1. Let V = jtr2h, where r ~ ro and h ~ ho. What re-
lationship must hold between ro and ho for V to be
equally sensitive to small changes in r and hi
2. (a) Find the unique critical point of the function
f(xux2, ...,xn) = c-*?-*l— -<
(b) Use the Hessian criterion to determine the nature
of this critical point.
3. The Java Joint Gourmet Coffee House sells top-of-
the-line Arabian Mocha and Hawaiian Kona beans.
If Mocha beans are priced at x dollars per pound
and Kona beans at y dollars per pound, then mar-
ket research has shown that each week approximately
80 — lOOx + 40y pounds of Mocha beans will be sold
and 20 + 60x — 35y pounds of Kona beans will be
sold. The wholesale cost to the Java Joint owners is
$2 per pound for Mocha beans and $4 per pound for
Kona beans. How should the owners price the coffee
beans in order to maximize their profits?
4. The Crispy Crunchy Cereal Company produces three
brands, X, Y, and Z, of breakfast cereal. Each month,
x, y, and z (respectively) 1000-box cases of brands X,
Y, and Z are sold at a selling price (per box) of each
cereal given as follows:
Brand
No. cases sold
Selling price per box
X
X
4.00 - 0.02x
Y
y
4.50-0.05y
Z
z
5.00-O.lOz
(a) What is the total revenue R if x cases of brand X, y
cases of brand Y, and z cases of brand Z are sold?
(b) Suppose that during the month of November, brand
X sells for $3.88 per box, brand Y for $4.25, and
brand Z for $4.60. If the price of each brand is in-
creased by $0. 10, what effect will this have on the
total revenue?
(c) What selling prices maximize the total revenue?
5. Find the maximum and minimum values of the
function
f(x, y, z) = x — V3 y
on the sphere x2 + y2 + z2 = 4 in two ways:
(a) by using a Lagrange multiplier;
Miscellaneous Exercises for Chapter 4
the rectangle without overlapping, except along their
edges. (See Figure 4.42.)
(b) by substituting spherical coordinates (thereby de-
scribing the point {x,y, z) on the sphere as x =
2 sin (p cos 9, y = 2 sirup sin#, z = 2cos<p) and
then finding the ordinary (i.e., unconstrained) ex-
trema of f(x(cp, 9), y(<p, 9), z((p, 9)).
6. Suppose that the temperature in a space is given by the
function
T(x, y,z) = 200xyz2.
Find the hottest point(s) on the unit sphere in two ways:
(a) by using Lagrange multipliers;
(b) by letting x = sirup cos 6*, y = sirup sin 9, z =
cos <p and maximizing T as a function of the two
independent variables <p and 9 . (Note: It will help if
you use appropriate trigonometric identities where
possible.)
7. Consider the function f(x, y) = (y — 2x2)(y — xz).
(a) Show that / has a single critical point at the origin.
(b) Show that this critical point is degenerate. Hence,
it will require means other than the Hessian crite-
rion to determine the nature of the critical point as
a local extremum.
(c) Show that, when restricted to any line that passes
through the origin, / has a minimum at (0, 0).
(That is, consider the function F(x) = f(x, mx),
where m is a constant and the function G(y) =
f(0, y).)
(d) However, show that, when restricted to the
parabola y = |jc2, the function / has a global
maximum at (0, 0). Thus, the origin must be a sad-
dle point.
(e) Use a computer to graph the surface z = f(x, y).
8. (a) Find all critical points of f(x, y) = xy that satisfy
x2 + y2 = 1.
(b) Draw a collection of level curves of / and, on the
same set of axes, the constraint curve x2 + y2 = 1 ,
and the critical points you found in part (a).
(c) Use the plot you obtained in part (b) and a geomet-
ric argument to determine the nature of the critical
points found in part (a).
9. (a) Find all critical points of f(x,y,z) = xy that
satisfy x2 + y2 + z2 = 1 .
(b) Give a rough sketch of a collection of level surfaces
of / and, on the same set of axes, the constraint
surface x2 + y2 + z2 = 1 , and the critical points
you found in part (a).
(c) Use part (b) and a geometric argument to determine
the nature of the critical points found in part (a).
10. Find the area A of the largest rectangle so that two
squares of total area 1 can be placed snugly inside
Figure 4.42 Figure for Exercise 10.
1 1 . Find the minimum value of
f{x\ ,x2, ... , x„) = x\ + x\ H h x2
subject to the constraint that a\X\ + a2x2 + • • • +
a„xn = 1, assuming that a\ + a\ + • • • + a2 > 0.
12. Find the maximum value of
f(xux2, ...,xn) = (aiXi + a2x2 H h anxn)2,
subject to x\ + x\ + • • • + x2 = I. Assume that not all
of the a, 's are zero.
1 3. Find the dimensions of the largest rectangular box that
can be inscribed in the ellipsoid x2 + 2y2 + 4z2 = 12.
Assume that the faces of the box are parallel to the
coordinate planes.
14. Your company must design a storage tank for Super
Suds liquid laundry detergent. The customer's specifi-
cations call for a cylindrical tank with hemispherical
ends (see Figure 4.43), and the tank is to hold 8000 gal
of detergent. Suppose that it costs twice as much (per
square foot of sheet metal used) to machine the hemi-
spherical ends of the tank as it does to make the cylin-
drical part. What radius and height do you recommend
for the cylindrical portion so as to minimize the total
cost of manufacturing the tank?
Figure 4.43 The storage tank
of Exercise 14.
15. Find the minimum distance from the origin to the
surface x2 — (y — z)2 = 1 .
1 6. Determine the dimensions of the largest cone that can
be inscribed in a sphere of radius a.
17. Find the dimensions of the largest rectangular box
(whose faces are parallel to the coordinate planes) that
Chapter 4 ! Maxima and Minima in Several Variables
can be inscribed in the tetrahedron having three faces in
the coordinate planes and fourth face in the plane with
equation bcx + acy + abz = abc, where a, b, and c
are positive constants. (See Figure 4.44.)
Figure 4.44 Figure for Exercise 17.
1 8. You seek to mail a poster to your friend as a gift. You
roll up the poster and put it in a cylindrical tube of di-
ameter a' and length y. The postal regulations demand
that the sum of the length of the tube plus its girth (i.e.,
the circumference of the tube) be at most 108 in.
(a) Use the method of Lagrange multipliers to find the
dimensions of the largest- volume tube that you can
mail.
(b) Use techniques from single-variable calculus to
solve this problem in another way.
1 9. Find the distance between the line y = 2x + 2 and the
parabola x = y2 by minimizing the distance between
a point (x\ , yi) on the line and a point (X2 , yi) on the
parabola. Draw a sketch indicating that you have found
the minimum value.
20. A ray of light travels at a constant speed in a uniform
medium, but in different media (such as air and water)
light travels at different speeds. For example, if a ray of
light passes from air to water, it is bent (or refracted)
as shown in Figure 4.45. Suppose the speed of light
A
\^ Medium
1
a i V
i V1!
h
i
i
Medium 2 10 £
i
B
in medium 1 is v\ and in medium 2 is 1)2 ■ Then, by
Fermat's principle of least time, the light will strike the
boundary between medium 1 and medium 2 at a point
P so that the total time the light travels is minimized.
(a) Determine the total time the light travels in going
from point A to point B via point P as shown in
Figure 4.45.
(b) Use the method of Lagrange multipliers to estab-
lish Snell's law of refraction: that the total travel
time is minimized when
sin 6*i
sin 6*2
Vl
!'2 '
(Hint: The horizontal and vertical separations of
A and B are constant.)
21. Use Lagrange multipliers to establish the formula
\ax0 + by0 — d\
D
V«2 + b2
for the distance D from the point (xq, yo) to the line
ax + by = d.
22. Use Lagrange multipliers to establish the formula
\ax0 + by0 + czo - d\
D
si a2 + b2 + c2
for the distance D from the point (xq, yo, zo) to the
plane ax + by + cz = d.
23. (a) Show that the maximum value of f(x,y,z)
2 2'
xAyAz
■2 subject to the constraint that x2 + y2 +
2 2
Z = a is
27 V 3
(b) Use part (a) to show that, for all x, y, and z,
(*Vz2)1/3
x2 + y2 + z2
(c) Show that, for any positive numbers x\ , x% , . . . , xn ,
i ^ xl+x2 + ---+xn
n
The quantity on the right of the inequality is the
arithmetic mean of the numbers x\, X2, ■ ■ . , xn,
and the quantity on the left is called the geomet-
ric mean. The inequality itself is, appropriately,
called the arithmetic-geometric inequality.
(d) Under what conditions will equality hold in the
arithmetic-geometric inequality?
In Exercises 24-27 you will explore how some ideas from ma-
trix algebra and the technique of Lagrange multipliers come
together to treat the problem of finding the points on the unit
hypersphere
Figure 4.45 Snell's law of refraction.
g(xu ■■■,x„)
+ x2 + -
l
4.6 | Miscellaneous Exercises for Chapter 4
that give extreme values of the quadratic form
f(*L
where the Ojj s are constants.
24. (a) Use a Lagrange multiplier X to set up a system of
n + 1 equations in w + 1 unknowns xi , . . . , xn , X
whose solutions provide the appropriate con-
strained critical points.
(b) Recall that formula (2) in §4.2 shows that the
quadratic form / may be written in terms of ma-
trices as
f(xi
,x„)
(1)
where the vector x is written as the « x 1 matrix
X\
and A is the n x n matrix whose ij th entry
is atj . Moreover, as noted in the discussion in §4.2,
the matrix A may be taken to be symmetric (i.e., so
that AT = A), and we will therefore assume that
A is symmetric. Show that the gradient equation
V/ = XV g is equivalent to the matrix equation
Ax = Xx.
(2)
Since the point (xi, . . . , xn) satisfies the constraint
*]+•••+ x% = 1, the vector x is nonzero. If you
have studied some linear algebra, you will recog-
nize that you have shown that a constrained criti-
cal point (xi, . . . , x„) for this problem corresponds
precisely to an eigenvector of the matrix A asso-
ciated with the eigenvalue X.
(c) Now suppose that x :
x\
is one of the eigen-
vectors of the symmetric matrix A, with associated
eigenvalue X. Use equations (1) and (2) to show, if
x is a unit vector, that
/(*!
■ , Xn) = X.
Hence, the (absolute) minimum value that /
attains on the unit hypersphere must be the small-
est eigenvalue of A and the (absolute) maximum
value must be the largest eigenvalue.
25. Let n = 2 in the situation of Exercise 24, so that we are
considering the problem of finding points on the circle
x2 + y2 = 1 that give extreme values of the function
f(x, y) = ax2 + 2bxy + cy2
[* y]
(a) Find the eigenvalues of A
a b
b c
by identify-
ing the constrained critical points of the optimiza-
tion problem described above.
(b) Now use some algebra to show that the eigenval-
ues you found in part (a) must be real. It is a fact
(that you need not demonstrate here) that any n x n
symmetric matrix always has real eigenvalues.
26. In Exercise 25 you noted that the eigenvalues X\, X2
that you obtained are both real.
(a) Under what conditions does X\ = X2I
(b) Suppose that X\ and X2 are both positive. Explain
why / must be positive on all points of the unit
circle.
(c) Suppose that Xi and X2 are both negative. Explain
why / must be negative on all points of the unit
circle.
27. Let / be a general quadratic form in n variables de-
termined by an n x n symmetric matrix A, that is,
f{x\, x„) = auxixj = xr^x-
(a) Show, for any real number k, that f(kx\,...,
kx„) = k2f(x\, . . . , x„). (This means that a
quadratic form is a homogeneous polynomial of
degree 2 — see Exercises 37-44 of the Miscella-
neous Exercises for Chapter 2 for more about ho-
mogeneous functions.)
(b) Use part (a) to show that if / has a positive
minimum on the unit hypersphere, then / must
be positive for all nonzero x e R" and that if /
has a negative maximum on the unit hypersphere,
then / must be negative for all nonzero x € R" .
(Hint: For x / 0, let u = x/||x||, so that x = ku,
where k = ||x||.)
(c) Recall from §4.2 that a quadratic form / is said
to be positive definite if /(x) > 0 for all nonzero
x e R" and negative definite if /(x) < 0 for all
nonzero x e R" . Use part (b) and Exercise 24 to
show that the quadratic form / is positive definite
if and only if all eigenvalues of A are positive, and
negative definite if and only if all eigenvalues of
A are negative. (Note: As remarked in part (b) of
Exercise 25, all the eigenvalues of A will be real.)
Multiple Integration
5.1 Introduction: Areas and
Volumes
5.2 Double Integrals
5.3 Changing the Order of
Integration
5.4 Triple Integrals
5.5 Change of Variables
5.6 Applications of Integration
5.7 Numerical Approximations
of Multiple Integrals
(optional)
True/False Exercises for
Chapter 5
Miscellaneous Exercises for
Chapter 5
y
Figure 5.1 The graph of
y = f(x).
y
Figure 5.2 The shaded region has
area /* f(x)dx.
5.1 Introduction: Areas and Volumes
Our purpose in this chapter is to find ways to generalize the notion of the definite
integral of a function of a single variable to the cases of functions of two or
three variables. We also explore how these multiple integrals may be used to
meaningfully represent various physical quantities.
Let / be a continuous function of one variable defined on the closed interval
[a, b] and suppose that / has only nonnegative values. Then the graph of / looks
like Figure 5.1. That / is continuous is reflected in the fact that the graph consists
of an unbroken curve. That / is nonnegative-valued means that this curve does not
dip below the x -axis. We know from one- variable calculus that the definite integral
fb f(x) dx exists and gives the area under the curve, as shown in Figure 5.2.
Now suppose that / is a continuous, nonnegative-valued function of two
variables defined on the closed rectangle
R = [(x, y) e R2 | a < x < b, c < y < d]
in R2. Then the graph of / over R looks like an unbroken surface that never dips
below the xy-plane, as shown in Figure 5.3. In analogy with the single-variable
case, there should be some sort of integral that represents the volume under the
part of the graph that lies over R. (See Figure 5.4.) We can find such an integral
by using Cavalieri's principle, which is nothing more than a fancy term for the
method of slicing. Suppose we slice by the vertical plane x = xo, where xo is a
constant between a and b. Let A(xo) denote the cross-sectional area of such a
slice. Then, roughly, one can think of the quantity A(xo) dx as giving the volume
of an "infinitely thin" slab of thickness dx and cross-sectional area A(xo). (See
Figure 5.5.) Hence, the definite integral
V
-f
J Li
A(x) dx
gives a "sum" of the volumes of such slabs and can be considered to provide a
reasonable definition of the total volume of the solid.
But what about the value of A(xo)? Note that A(xo) is nothing more than the
area under the curve z = f(xo, y), obtained by slicing the surface z = f(x, y)
with the plane x = xq. Therefore,
A(x0)
f(x0, y)dy
5.1 | Introduction: Areas and Volumes 311
Figure 5.3 The graph of Figure 5.4 The region under the Figure 5.5 A slab of "volume"
z = f(x, v). portion of the graph of / lying dV = A(x0) dx.
over R has volume that is given
by an integral.
Plane z = c
(a, 0, 0)
(0,6,0)
y
(a, b, 0)
Figure 5.6 Calculating the
volume of the box of Example 1 .
Figure 5.7 The graph of
z = 4 - x2
Example 2.
4 - x2 - y2 of
(remember xq is a constant), and so we find that
o pupa
= / A(x)(ix = / / /(x, y)dy
J a J a lJ c
dx.
(i)
The right-hand side of formula (1) is called an iterated integral. To calculate
it, first find an "antiderivative" of f(x, y) with respect to y (by treating i as a
constant), evaluate at the integration limits y = c and y = d, and then repeat the
process with respect to x.
EXAMPLE 1 Let's make sure that the iterated integral defined in formula (1)
gives the correct answer in a case we know well, namely, the case of a box. We'll
picture the box as in Figure 5.6. That is, the box is bounded on top and bottom by
the planes z = c (where c > 0) and z = 0, on left and right by the planes y = 0
and y = b (where b > 0), and on back and front by the planes x = 0 and x = a
(a > 0). Hence, the volume of the box may be found by computing the volume
under the graph of z = c over the rectangle
R = {(x,y) | 0 < x < a,0 < y < b}.
Using formula (1), we obtain
V = y j cdydx = j (^cy\-y~^ dx = j cbdx = cbx\]zCQ = cba.
This result checks with what we already know the volume to be, as it should. ♦
EXAMPLE 2 We calculate the volume under the graph of z = 4 — x2 — y2
(Figure 5.7) over the square
R = {(x, y) | -1 < x < 1, -1 < y < 1}.
Using formula (1) once again, we calculate the volume by first integrating with
respect to y (i.e., by treating x as a constant in the inside integral) and then by
Chapter 5 | Multiple Integration
y = y0 plane
gure 5.8
first.
Y/y
Slicing by y = yo
integrating w
■/.
th respect to x. The details are as follows:
j (4-x2- y2)dydx = j
dx
y=-l
A-xz
8 - 2x2
22 2 :
X X
3 3
-4 + x2 +
dx
dx
-l
22
T
22 2
T + 3
40
y
In our development of formula (1 ), we could just as well have begun by slicing
the solid with the plane y = yo (instead of with the plane x = xq), as shown in
Figure 5.8. Then, in place of formula (1), the formula that results is
V
rd rb
= I I f(x, y )dxdy.
J c J a
(2)
Since the iterated integrals in formulas ( 1 ) and (2) both represent the volume of the
same geometric object, we can summarize the preceding discussion as follows.
PROPOSITION 1 .1 Let R be the rectangle {(x, y) \ a < x < b, c < y < d) and
let / be continuous and nonnegative on R. Then the volume V under the graph
of /' over R is
pb pel pel pb
/ / f(x,y)dydx= / f(x,y)dxdy.
J a J c J c J a
EXAMPLE 3 We find the volume under the graph of z = cosx siny over the
rectangle
R
f 7T TV 1
= \(x,y) \ 0<x <-, 0<y <-\
(See Figure 5.9.) From formula (1), we calculate that the volume is
V
fJt/2 fx/2
= / cos x sin y dy dx = /
Jo Jo Jo
*'2 ( sj2_
2
(— cosx cos j)|y=o/4 dx
2-V2 r12
— 2~ — J
2- s/2
sinx
7T/2
2 - V2 2 - s/2
—^(1-0) = —^-.
5.1 | Exercises 313
Figure 5.9 The surface z = cosx siny of
Example 3.
If we use formula (2) instead of formula (1), we obtain
tjt/4 fji/2 rir/4
Jo Jo
I cos x sin y dx dy = / (sinx sin y)\xx=^2 dy
Jo Jo
*n/4
= f
Jo
,t/4
(sin y — 0) dy = — cos y |0
4l 2-yfl
= ~- (-1) = — •
2 V ' 2
That this result agrees with our first calculation is no surprise given
Proposition 1.1. ♦
5.1 Exercises
Evaluate the iterated integrals given in Exercises 1-6.
2. J J y sin x dy dx
3. / / xey dy dx
J -2 Jo
rn/2 ri
4. / / ex cos y dx dy
Jo Jo
5. J j (ex+y + x2 + In y) dx dy
6. f f — ^— dx dy
Jl Ji xy
7. Find the volume of the region that lies under the graph
of the paraboloid z = x2 + y2 +2 and over the rect-
angle R = {(x, y) | — 1 < x < 2, 0 < y < 2} in two
ways:
(a) by using Cavalieri's principle to write the volume
as an iterated integral that results from slicing the
region by parallel planes of the form x = constant;
(b) by using Cavalieri's principle to write the volume
as an iterated integral that results from slicing the
region by parallel planes of the form y = constant.
8. Find the volume of the region bounded on top by the
plane z = x + 3y + 1, on the bottom by the xy-plane,
and on the sides by the planes x = 0, x = 3, y = 1,
y = 2.
9. Find the volume of the region bounded by the graph
of f{x, y) = 2x2 + y4 sinjrx, the xy-plane, and the
planes x = 0, x = 1, y = — 1, y = 2.
314 Chapter 5 | Multiple Integration
In Exercises 10—15, calculate the given iterated integrals and
indicate of what regions in R3 they may be considered to
represent the volumes.
10. j j Idxdy
11. J j (16-x2 -y2)dydx
/jr/2 />jt
I sin x cos y dx dy
-jr/2 JO
13. j j (4-x2)dxdy
14. J J \x\ sin jry dy dx
15. J J (5-\y\)dxdy
16. Suppose that / is a nonnegative- valued, continu-
ous function defined on R = {(x, y) | a < x < b, c <
y < d}. If fix, y) < M for some positive number M,
explain why the volume V under the graph of / over
R is at most M(b — a)(d — c).
5.2 Double Integrals
In the previous section we saw how to calculate volumes of certain solids us-
ing iterated integrals. The ideas were mostly straightforward, but the situation
we addressed was rather special: We only solved the problem of computing the
volume of a solid defined as the region lying under the graph of a continuous,
nonnegative-valued function f(x, y) and above a rectangle in the xv-plane. It is
not immediately apparent how we might compute the volume of a more general
solid based on this work.
Thus, in this section we define a more general notion of an integral of a
function of two variables that will allow us to describe
1. integrals of arbitrary functions (i.e., functions that are not necessarily non-
negative or continuous) and
2. integrals over arbitrary regions in the plane (i.e., rather than integrals over
rectangles only).
We focus first on case 1 . To do this, we start fresh with some careful definitions
and notation. The ideas involved in Definitions 2.1-2.3 below are different from
those in the previous section. However, we will see that there is a key connection
(called Fubini's theorem) between the notion of an iterated integral discussed
in §5.1 and that of a double integral, which will be described in Definition 2.3.
The Integral over a Rectangle
We also denote a (closed) rectangle
R = {{x, y) e R2 | a < x < b, c < y < d}
by [a, b] x [c, d]. This notation is intended to be analogous to the notation for a
closed interval.
DEFINITION 2.1 Given a closed rectangle R = [a, b] x [c, d], a partition
of R of order n consists of two collections of partition points that break up
R into a union of n2 subrectangles. More specifically, for i, j = 0, . . . , n, we
introduce the collections {*, } and {yj}, so that
a = xq < x\ < ■ ■ ■ < x,_i < Xj < ■ ■ ■ < x„ = b,
5.2 | Double Integrals 315
and
c = yo < yi <
<
yj-y < yj
< yn = d.
Let Ax,- = x,- — Xi-i (for i = 1, . . . , n) and Ay;- = y,- — y;_i (for j =
1, . . . , n). Note that Ax,- and Ayv- are just the width and height (respec-
tively) of the z'y'th subrectangle (reading left to right and bottom to top) of the
partition.
An example of a partitioned rectangle is shown in Figure 5.10. We do not
assume that the partition is regular (i.e., that all the subrectangles have the same
dimensions).
d = y„
yj-i
c=y0
+
H h
+
+
+
CI — Xq X-[ X2 ' ■ ■ Xj _ 1 Xj • • • Xn — D
Figure 5.10 A partition of the rectangle [a, b] x [c, d].
DEFINITION 2.2 Suppose that / is any function denned on R = [a, b] x
[c, d] and partition R in some way. Let c,;- be any point in the subrectangle
Rij = [xi-\,Xi\ x [yj-uyj] (i,j = l,..., n).
Then the quantity
n
where AAjj = Ax, Ay7 is the area of Rij, is called a Riemann sum of / on
R corresponding to the partition.
The Riemann sum
S = £/(cy)AA(7
ij
depends on the function /, the choice of partition, and the choice of the "test
point" cy in each subrectangle Rij of the partition. The Riemann sum itself is
just a weighted sum of areas AAjj of subrectangles of the original rectangle R,
the weighting being given by the value /(cy).
If / happens to be nonnegative on R, then, for i, j = 1 , . . . , n, the individual
terms /(cy ) A Ay in S may be considered to be volumes of boxes having base area
Chapter 5 | Multiple Integration
z
Figure 5.11 The volume under the graph Figure 5.12 The Riemann sum as a signed sum of
of / is approximated by the Riemann sum. volumes of boxes.
Ax, Ay j and height /(cy ). Therefore, S can be considered to be an approximation
to the volume under the graph of / over R, as suggested by Figure 5.1 1. If / is
not necessarily nonnegative, then the Riemann sum S is a signed sum of such
volumes (because, with /(cy) < 0, the term /(Cij)AA,j is the negative of the
volume of the appropriate box — see Figure 5.12).
Al\
Figure 5.13 HAUA2,A3
represent the values of the shaded
areas, then
f* f(x)dx = Al-A2 + A3.
R
(in xy-plane)
Figure 5.14 If V\, V2 represent
the volumes of the shaded regions,
then ffR f(x, y)d A = Vl-V2.
DEFINITION 2.3 The double integral of / on R, denoted by ffRf dA
(or by f fR f{x, y) dA or by / fR f(x, y) dx dy), is the limit of the Riemann
sum S as the dimensions Ax; and Ayj of the subrectangles Rjj all approach
zero, that is,
f f fdA = lim V f(Cij)AxiAyj,
provided of course, that this limit exists. When ffRf dA exists, we say that
/ is integrable on R.
The crucial idea to remember — indeed, the defining idea — is that the integral
ffRf dA is a limit of Riemann sums S, for this concept is what is needed to
properly apply double integrals to physical situations.
From a geometric point of view, just as the single-variable definite integral
fa f(x) dx can be used to compute the "net area" under the graph of the curve
v = f(x) (as in Figure 5. 13), the double integral ffR fdA can be used to compute
the "net volume" under the graph of z = fix, y) (as in Figure 5.14).
Another way to view the double integral ffRf dA is somewhat less geometric
but is more in keeping with the notion of the integral as the limit of Riemann
sums and provides a perspective that generalizes to triple integrals of functions of
three variables. Instead of visualizing the graph of z = fix, y) as a surface and
S = Yll j=\ f(cij)AxiAyj as a (signed) sum of volumes of boxes related to the
graph, consider S to be a weighted sum of areas and the integral ffRf dA the
limiting value of such weighted sums as the dimensions of all the subrectangles
approach zero. With this point of view, we do not depict the integrand / when we
try to visualize the integral. In this way, the distinction between the roles of the
integrand and the rectangle R over which we integrate can be made clearer. (See
Figure 5.15.)
5.2 | Double Integrals 317
R
- This subrectangle of
area A^453 contributes
/(c53) AA53 to S.
Figure 5.15 S = /(Cy)AAy.
EXAMPLE 1 Suppose that a 3 cm square metal plate is made, but some nonuni-
formities exist due to the manufacturing process so that the mass density varies
somewhat throughout the plate. If we knew the density function 8(x, y) at every
point in the plate, then we could calculate the total mass of the plate as
Total mass = j j S(x,y)dA,
where D denotes the square region of the plate placed in an appropriate coordinate
system.
In the absence of an analytic expression for S , we nonetheless can approximate
the double integral by means of a Riemann sum: We partition the square region
of the plate, take density readings at a test point in each subregion, and combine
to approximate the integral for the total mass. (Essentially what we are doing is
assuming that the density is nearly constant on each subregion so that multiplying
density and area will give the approximate mass of the subregion; adding these
approximate masses then gives an approximation for the total mass.) For example,
we might model the problem as in Figure 5.16, where the region of the plate is
(0.6)
(0.3)
(0.2)
(0.2)
(0.1)
(0.5)
(0.3)
(0.3)
Figure 5.16 The region of Example 1. The 3x3
square is partitioned into nine subregions. The density
values at test points in each subregion are shown.
Chapter 5 | Multiple Integration
Figure 5.17 The graph
of z = x of Example 2.
HI
Figure 5.1 8 The two
subrectangles R^j and Ri2j are
symmetrically placed with respect
to the y-axis. The corresponding
test points C;, / and c>2 j are chosen
so that they have the same
y-coordinates and opposite
x-coordinates.
partitioned into nine square subregions. Then we have
Total mass = j j S(x, y)dA «a ^ 5(c,7)AAy
= (0.2)1 + (0.3)1 + (0.6)1 + (0.1)1 + (0.2)1 + (1)1 + (0.3)1
+ (0.3)1 +(0.5)1 = 3.5. ^
EXAMPLE 2 We determine the value of ffRxdA, where R = [-2,2] x
[—1, 3]. Here the integrand f(x, y) = x and, if we graph z = f(x, y) over R,
we see that we have a portion of a plane, as shown in Figure 5.17. Note that the
portion of the plane is positioned so that exactly half of it lies above the xy-plane
and half below. Thus, if we regard ffRx dA as the net volume under the graph of
z = x, then we conclude that ffRx dA (if it exists) must be zero.
On the other hand, we need not resort to visualization in three dimensions.
Consider a Riemann sum corresponding to ffRx dA obtained by partitioning
R = [—2, 2] x [—1,3] symmetrically with respect to the y-axis and by choosing
the "test points" c,; symmetrically also. (See Figure 5.18.) It follows that the value
of
S = f(cu)AAij = J2XiiAAiJ
(where x,7 denotes the x-coordinate of cy) must be zero since the terms of the
sum cancel in pairs. Furthermore, we can arrange things so that, as we shrink
the dimensions of the subrectangles to zero (as we must do to get at the integral
itself), we preserve all the symmetry just described. Hence, the limit under these
restrictions will be zero, and thus, the overall limit (where we do not impose such
symmetry restrictions on the Riemann sum), if it exists at all, must be zero as
well. ♦
Example 2 points out fundamental difficulties with Definition 2.3, namely,
that we never did determine whether ffRf dA really exists. To do this, we would
have to be able to calculate the limit of Riemann sums of / over all possible
partitions of R by using all possible choices for the test points c,y, a practically
impossible task. Fortunately, the following result (which we will not prove) pro-
vides an easy criterion for integrability:
Figure 5.1 9 The graph of a
piecewise continuous function.
THEOREM 2.4 If / is continuous on the closed rectangle R, then ffR f dA
exists.
In Example 2, f(x, y) = x is a continuous function and hence integrable
by Theorem 2.4. The symmetry arguments used in the example then show that
ffRxdA = 0,
Continuous functions are not the only examples of integrable functions. In
the case of a function of a single variable, piecewise continuous functions are also
integrable. (Recall that a function f(x) is piecewise continuous on the closed
interval [a, b] if / is bounded on [a, b] and has at most finitely many points of
discontinuity on the interior of [a , b] . Its graph, therefore, consists of finitely many
continuous "chunks" as shown in Figure 5.19.) For a function of two variables,
there is the following result, which generalizes Theorem 2.4.
5.2 | Double Integrals 319
THEOREM 2.5 If / is bounded on R and if the set of discontinuities of / on R
has zero area, then ffR f dA exists.
To say that a set X has zero area as we do in Theorem 2.5, we mean that we
can cover X with rectangles R\, R2, . . . , Rni . . . (i.e., so that X c [J^Lj ^«)> tne
sum of whose areas can be made arbitrarily small.
A function / satisfying the hypotheses of Theorem 2.5 has a graph that
looks roughly like the one in Figure 5.20. Theorem 2.5 is the most general suffi-
cient condition for integrability that we will consider. It is of particular use to us
when we define the double integral of a function over an arbitrary region in the
plane.
a a
X
Discontinuities of/
Figure 5.20 The graph of an integrable
function.
Although Theorems 2.4 and 2.5 make it relatively straightforward to check
that a given integral exists, they do little to help provide the numerical value of
the integral. To mechanize the evaluation of double integrals, we will use the
following result:
THEOREM 2.6 (Fubini's theorem) Let / be bounded on R = [a, b] x [c, d]
and assume that the set S of discontinuities of / on R has zero area. If every line
parallel to the coordinate axes meets S in at most finitely many points, then
p p pb pd pd pb
// fdA= / / f(x,y)dydx= / / f(x,y)dxdy.
J J R J a J c J c J a
Fubini's theorem demonstrates that under certain assumptions the double in-
tegral over a rectangle (i.e., the limit of Riemann sums) can be calculated by using
iterated integrals and, moreover, that the order of integration for the iterated inte-
gral does not matter. We remark that the independence of the order of integration
depends strongly on the fact that the region of integration is rectangular; it will not
320 Chapter 5 | Multiple Integration
generalize to more arbitrary regions in such a simple way. (A proof of Theorem
2.6 is given in the addendum to this section.)
EXAMPLE 3 Werevisit ffRx dA in Example 2, where R = [-2, 2] x [-1, 3].
By Theorem 2.6, we know that JfRxdA exists and by Fubini's theorem, we
calculate
j j xdA = j j x dy dx = j
xy
=-i
dx
-L
x(3 - {-\))dx
/:
Ax dx = 2x
2|2
1-2
8 = 0,
which checks. Furthermore, we also have
j j x dA = j j xdxdy = j
J JR J-l J-2 J-]
dy = f_l(2-2)dy = 0.
x=—2
PROPOSITION 2.7 (Properties of the integral) Suppose that / and g are
both integrable on the closed rectangle R. Then the following properties hold:
1. / + g is also integrable on R and
/ / (f + g)dA= f f fdA+ f f gdA.
J Jr J J r J J r
2. cf is also integrable on R, where c e R is any constant, and
\LcfiA = c\lSdA'
3. If f(x, y) < g(x, y) for all (x,y)e R, then
If f{x,y)dA< I j g(x,y)dA.
4. | f\ is also integrable on R and
f Ir f Jr
< / / \f\dA.
Properties 1 and 2 are called the linearity properties of the double integral.
They can be proved by considering the appropriate Riemann sums and taking
limits. For example, to prove property 1, note that the Riemann sum whose limit
5.2 | Double Integrals 321
is
ffR(f + g)dA
is
Figure 5.21
in the plane.
A bounded region D
Figure 5.22 A type 1 elementary
region.
)
>
y = d
POO
y = c
Figure 5.23 A type 2 elementary
region.
t'J=i
= E /(cv)AA!7 + E *(c/y)AAy
U I 1,7 = 1
Property 3 (known as monotonicity) and property 4 can also be proved using
Riemann sums. For property 4, one needs to use the fact that
E«*
k=l
<E|a*!-
fc=l
Double Integrals over General Regions in the Plane
Our next step is to understand how to define the integral of a function over an
arbitrary bounded region D in the plane. Ideally, we would like to give a precise
definition of ffDfdA, where D is the amoeba-shaped blob shown in Figure 5.21
and where / is bounded on D. In keeping with the definition of the integral over
a rectangle, ffD f dA should be a limit of some type of Riemann sum and should
represent the net volume under the graph of / over D. Unfortunately, the techni-
calities involved in making such a direct approach work are prohibitive. Instead, we
shall consider only certain special regions (rather than entirely arbitrary ones), and
we shall assume that the integrand / is continuous over the region of integration
(which will allow us to use what we already know about integrals over rectangles).
Although this approach will not provide us with a completely general definition,
it is sufficient for essentially all the practical situations we will encounter.
To begin, we define the types of elementary regions we wish to consider.
DEFINITION 2.8 We say that D is an elementary region in the plane if it
can be described as a subset of R2 of one of the following three types:
Type 1 (see Figure 5.22):
D = {(*, y) | y(x) <y< S(x), a<x <b},
where y and <5 are continuous on [a, b].
Type 2 (see Figure 5.23):
D = {(x, y) | a(y) < x < 0(y), c<y<d},
where a and /3 are continuous on [c, d].
Type 3 D is of both type 1 and type 2.
Thus, a type 1 elementary region D has a boundary (denoted 3D) consisting
of straight segments (possibly single points) on the left and on the right and graphs
of continuous functions of x on the top and on the bottom. A type 2 elementary
322 Chapter 5 | Multiple Integration
region has a boundary that is straight on the top and bottom and consists of graphs
of continuous functions of y on the left and right.
EXAMPLE 4 The unit disk, shown in Figure 5.24, is an example of a type 3
elementary region. It is a type 1 region since
D = Ux,y) | — v/l - x2
y < yi — .
(See Figure 5.25.) It is also a type 2 region since
D = Ux, y) | —/l - y2 <x<
(See Figure 5.26.)
y
-1 <x < lj .
i<y<i}-
Figure 5.24 The unit disk
D = {(x, y) | a2 + v2 < 1} is a
type 3 region.
y = l
/
y = -l
Figure 5.25 The unit disk D as a
type 1 region.
Figure 5.26 The unit disk D as a type 2
region.
Now we are ready to define ffDf dA, where D is an elementary region and
/ is continuous on D. We construct a new function /ext, the extension of /, by
fex\x,y) =
f(x,y) if (x,y)eD
0 if(x,y)£D
Note that, in general, /ext will not be continuous, but the discontinuities of /ext
will all be contained in 3D, which has no area. Hence, by Theorem 2.5, /ext is
integrable on any closed rectangle R that contains D. (See Figure 5.27.)
Figure 5.27 The graph of z = /ext(jc, y).
5.2 | Double Integrals
jy = 3x2
Djl(l,3)
/ \ y = 4 - X2
(2,0)
Figure 5.28 The domain of / of
Example 5.
DEFINITION 2.9 Under the previous assumptions and notation, if R is any
rectangle that contains D, we define
fLfdA ,obe //«
/extJA.
Note that Definition 2.9 implicitly assumes that the choice of the rectangle R
that contains D does not affect the value of f fR fext dA. This is almost obvious
but still should be proved. We shall not do so directly but instead establish the
following key result:
THEOREM 2.10 Let D be an elementary region in R2 and / a continuous
function on D.
1. If D is of type 1 (as described in Definition 2.8), then
/ / fdA = / / f(x,y)dydx.
J J D Ja Jy(x)
2. If D is of type 2, then
r r rd pP(y)
fdA= / f(x,y)dxdy.
J J D Jc Ja(y)
Theorem 2.10 provides an explicit and straightforward way to evaluate double
integrals over elementary regions using iterated integrals. Before we prove the
theorem, let us illustrate its use.
EXAMPLE 5 Let D be the region bounded by the parabolas y = 3x2, y =
4 — x2 and the y-axis as shown in Figure 5.28. (Note that the parabolas intersect at
thepoint(l, 3).) Since D isatype 1 elementary region, we may use Theorem 2. 10
with f(x, y) — x2y to find that
2 fl r4^2 2
x ydA= II x ydydx.
Jo J\k2
The limits for the first (inside) integration come from the y-values of the top and
bottom boundary curves of D. The limits for second (outside) integration are the
constant x -values that correspond to the straight left and right sides of D. The
evaluation itself is fairly mechanical:
=4-x2
I I x2 ydydx = I I
J0 Jxx2 Jo V
1/22
' x y
dx
=3x2
= / y((4-*2)2-(3*2)>*
l fl
= -J x2 (16 - 8;t2 + x4 - 9x4) dx
= f (8.t2 - 4x4 - 4x6) dx =
Jo
8 4 4 _ 136
3 5 7 105-
324 Chapter 5 | Multiple Integration
Note that after the y -integration and evaluation, what remains is a single definite
integral in x. The result of calculating this x -integral is, of course, a number. Such
a situation where the number of variables appearing in the integral decreases with
each integration should always be the case. ♦
Proof of Theorem 2.10 For part 1, we may take D to be described as
D = {(x, y) | y(x) < y < S(x), a < x <b}.
We have, by Definition 2.9, that
fLfdA=fLf~dA-
where R is any rectangle containing D. Let R = [a', b'] x [c', d'], where a' < a,
b' > b, and c' < y(x), d' > S(x) for all x in [a, b]. That is, we have the situation
depicted in Figure 5.29. Since /ext is zero outside of the subrectangle R2 =
[a,b] x [c',d'],
f f fextdA= f f fextdA= f f fex\x,y)dydx
by Fubini's theorem. For a fixed value of x between a and b, consider the y-
integral fext(x, y)dy. Since fext(x, y) = 0 unless y(x) < y < S(x) (in which
case fext(x, y) = f(x, y)),
pel' pS(x)
/ rx\x,y)dy= / f(x,y)dy,
Jd Jy(x)
and so
f f f(x,y)dA= f f fxtdA= f f fex\x,y)dydx
J JD J J R Ja Jc'
/ / f(x,y)dydx,
as desired.
The proof of part 2 is very similar.
Figure 5.29 The region R is the union of Ri, R2,
and^?3 .
(0,1)'
>v x + y = 1
D
(0,0)
\(1.0)
Figure
5.30 The region D of
Example 6.
;
>
\ J = 1 - X |
x = 0
D
y = 0
Figure
5.31 The region D of
Example 6 as a type 1 region.
;
>
y = l
x = 0
\x = 1 — y
D \.
y = 0
Figure
5.32 The region D of
Example 6 as a type 2 region.
5.2 | Double Integrals 325
We continue analyzing examples of double integral calculations.
EXAMPLE 6 Let D be the region shown in Figure 5.30 having a triangular
border. Consider f fD(\ — x — y)dA. Note that D is a type 3 elementary region,
so there should be two ways to evaluate the double integral.
Considering D as a type 1 elementary region (see Figure 5.3 1), we may apply
part 1 of Theorem 2.10 so that
/ / (1 — x — y)dA= I I (1 — x — y)dydx
J 3d Jo Jo
-L
-L
-I
y-xy
y=\-x
dx
y=0
' 7 (1 -x)2\
(1 — x) — x(l — x) I dx
(1 - xf
dx= -\{\-x)\ = \.
We can also consider D as a type 2 elementary region, as shown in Figure 5.32.
Then, using part 2 of Theorem 2.10, we obtain
/ / (1 — x — y)dA = I I (l-x-y)dxdy.
J J D Jo Jo
We leave it to you to check explicitly that this iterated integral also has a value of
t . Instead we note that
Jo Jo
y ) dy dx
can be transformed into
/ / (1 — x — y) dx dy
Jo Jo
by exchanging the roles of x and y. Hence, the two integrals must have the same
value. In any case, the double integral
(1 — x — y)dA
represents the volume under the graph of z = 1 — x — y over the triangular region
D . If we picture the situation in R3 , as in Figure 5 . 3 3 , we see that the double integral
represents the volume of a tetrahedron. ♦
Of course, not all regions in the plane are elementary, including even some
relatively simple ones. To integrate continuous functions over such regions, the
best advice is to attempt to subdivide the region into finitely many of elementary
type.
326 Chapter 5 | Multiple Integration
EXAMPLE 7 Let D be the annular region between the two concentric circles
of radii 1 and 2 shown in Figure 5.34. Then D is not an elementary region, but we
can break D up into four subregions that are of elementary type. (See Figure 5.35.)
If f(x, y) is any function of two variables that is continuous (hence integrable)
on D, then we may compute the double integral as the sum of the integrals over
the subregions. That is,
[[ fdA=[[ fdA+ff fdA+ff fdA+[[ fdA.
J J D J J Di J JD2 J J D] J J D4
For the type 1 subregions, we have the set-up shown in Figure 5.36:
f(x, y)dy dx
and
\LSiA
V3 ,-1
f(x, y)dy dx.
-VI J—jA^?
For the type 2 subregions, we use the set-up shown in Figure 5.37:
f fo2 f-i f*/i
y = — V4 - x-
Figure 5.36 The subregions D\ and
Z?3 of Example 7 are of type 1 .
fix, y)dx dy
J
1
.2 _
\ ✓
\
\
s
X
✓
x= VT^y2
x = V4-y2
Figure 5.37 The subregions D2 and D4 of
Example 7 are of type 2.
5.2 | Double Integrals
and
JL<«-U.
<l-y2
-1 J-J4-V2
f(x, y)dx dy.
The difficulty of evaluating each of the preceding four iterated integrals then
depends on the complexity of the integrand. ♦
2
/ D
/x-y=0
~K ~l /
1 2
-l
Figure 5.38 The region D
of Example 8.
EXAMPLE 8 We calculate ffD ydA, where D is the region bounded by the
line x — y = 0 and the parabola x = y2 — 2. (See Figure 5.38.)
In this case D is a type 2 elementary region, where the left and right boundary
curves may be expressed as x = y2 — 2 and x = y, respectively. These curves
intersect where
y = -l,2.
y*-2 = y y^-y-2 = Q <
Therefore, part 2 of Theorem 2.10 applies to give
-2 py
1L
ydA
/I y dx dy
■1 Jv2-2
f i f2 2 xy\^_2 dy = j y- (y2 - 2)y) dy
j y-y'+2y) dy-
3 4 \ ^
y--y- + y2
3 4 y
= (f-4 + 4)-(-
3 4
+ !) = !■
Now, although D is not a type 1 elementary region, it may be divided into
two type 1 subregions along the vertical line x = — 1. (See Figure 5.39.) The
subregion D\ lying left of the line x = — 1 is bounded on both top and bottom by
the parabola x = y2 — 2; by solving for y we may express the bottom boundary
of D\ as y = —*Jx + 2 and the top boundary as y = \fx + 2. The subregion
D2 lying right of x = — 1 is bounded on the bottom by y = x and on the top by
Figure 5.39 The region D of Example 8 is divided
into subregions D\ and D2 by the line x = — 1.
328 Chapter 5 | Multiple Integration
y = -J x + 2. Putting all this information together, we have
/ / ydA = I I ydA+ f f ydA
/ — 1 p*/x+2 r2 r
I _ydydx+ / /
-2 J-Jx+2 J -I Jx
y dy dx
-1 y2
<-2 ~2
-1
y=Vx+2
dx +
£
2 2 y=vx+2
y=-Vx+2
/■2/x + 2 *2\
- y ( —
, x3
1 4+*"T
= (l+2-l)-(i-l + i)
4'
Addendum: Proof of Theorem 2.6
Step 1. First we establish Theorem 2.6 in the case where / is continuous on
R = [a, b] x [c, d]. By Theorem 2.4, we know that ffRf dA exists. Let F be
the single-variable function denned by
F(x)
f(x,y)dy.
(Note: Since / is continuous on R, the partial function in y obtained by holding
x constant is continuous on [c, d]. Hence, fd fix, y)dy exists for every x in
[a, b].) We show that
nb nb r nd n n
/ F(x)dx= / / f(x,y)dy dx = f dA.
J a J a \_Jc J J JR
Let a = xq < x\ < • • • < x„ = b be any partition of [a, b]. Then a general
Riemann sum that approximates F(x) dx is
//
j>(jtf)A*,, (1)
where Ax; = x;- — andx* e [x,-_i, x,].Nowletc = y0 < < • • • < = d
be a partition of [c, d]. (The partitions of [a, b] and [c, J] together give a partition
of R = [a, b] x [c, rf].) Therefore, we may write
x) = Jdf(x,
y)dy
fix, y)dy +
y)dy +
t f
7=1 Jy>-1
f(x,y)dy.
+ f fix,y)dy
Jy„-i
5.2 | Double Integrals 329
By the mean value theorem for integrals,1 on each subinterval [yj-i, yj] there
exists a number y* such that
f
Jy,
f(x, y)dy = (yj - y}-i)f(x, y*) = f(x, y*)Ayj.
The choice of y* in general depends on x, so henceforth, we will write y*(x) for
y*. Consequently,
n
/•(.v)- ^./(.v.y;(.v)).\y,.
and the Riemann sum (1) may be written as
i=l I 7=1
Axi = /(C;,).\.V,.\.V(.
where cy = (xf , y*(xf)). Note that Cy e x;] x [y;_i, y,-]. (See Fig-
ure 5.40.)
y;(*,*)
c •
H 1 — h
H h
1 « I
Figure 5.40 The point Cy = (.«*, y *(*,*)) used in
the proof of Theorem 2.6.
We have thus shown that given any partition of [a,b], we can associate a
suitable partition of R = [a, b] x [c, d] such that the Riemann sum (1) that ap-
proximates fa F(x)dx is equal to a Riemann sum (namely, J2i j f(cij)Axj Ayj)
that approximates ffRf dA. Since / is continuous, we know that
/(cy)Ajc,- Ay7- approaches / / f dA
U J Js
as Ax, and Ay,- tend to zero. Hence,
I F(x)dx = 11 fdA.
Ja J JR
By exchanging the roles of x and y in the foregoing argument, we can show
that
II fdA = f I f(x,y)dxdy.
J J R Jc Ja
1 The mean value theorem for integrals says that, if g is continuous on [a, b], then there is some number
c with a < c < b such that f g(x)dx = (b — a)g(c).
Chapter 5 | Multiple Integration
Step 2. Now we prove the general case of Theorem 2.6 (i.e., the case that
/ has discontinuities in R = [a, b] x [c, d]). By hypothesis, the set S of discon-
tinuities of / in R are such that every vertical line meets S in at most finitely
many points. Thus, for each x in [a, b], the partial function in y of f(x, y) is
continuous throughout [c, d], except possibly at finitely many points. (In other
words, the partial function is piecewise continuous.) Then, because / is bounded,
(x) = ^ f(x
F(x)= I f(x,y)dy
exists.
Now we proceed as in Step 1 . That is, we begin with a partition of [a, b] into
n subintervals and a corresponding Riemann sum
i=i
Next, we partition [c, d] into n subintervals. Hence,
Fix*) = f f(x*, y)dy = J2 f' fix*, y)dy. (2)
As in Step 1, the partitions of [a, b] and [c, d] combine to give a partition of R.
Write R as U R2, where R\ is the union of all subrectangles
Rij = [Xi-uXf] x [y;_!, yj]
that intersect S and R2 is the union of the remaining subrectangles. Then we may
apply the mean value theorem for integrals to those intervals [yj-i, yj] on which
fix*, y) is continuous in y, thus replacing the integral
/ f(xf,y)dy
by
f(x*,y*(x*))Ayj = f(cij)Ayj.
Since / is bounded, we know that
\f(x,y)\<M
for some M and all (x, y) e R. Therefore, on the intervals [y7_i, yj] where
fix* , y) fails to be continuous, we have
f1 fix*,y)dy < r \ f(x*,y)\dy
< r Mdy = M(yj - y;_j) = MAyr (3)
Jyj-i
From equation (2), we know that
f2Hx*)Axi = it,\f} fix*,y)dy}AXi
= E f fix*,y)dy\AXi
5.2 | Double Integrals 331
f(x*,y)dy\Axi
f(x*,y)dy\Ax
Therefore,
= s \i:
A',,t K- 1 *' '• •
J>(x*)Ax,- J] { r f(xf,y)dy\Axi
- (f
f(x*,y)dy^Axi
(4)
Applying the mean value theorem for integrals to the left side of equation (4) and
inequality (3) to the right side, we obtain
J2nx*)Axi- f(Cij)AxtAyj
i = l RijCRi
MAxjAyj
RijCRi
M ■ area of R\ .
Now S has zero area (by hypothesis) and is contained in R\. By letting the
partition of R become sufficiently fine (i.e., by making Ax, , Ay, small), the term
M ■ area of R\ can be made arbitrarily small. (See Figure 5.41.)
d -
c -
- —
N
<
f
i
— -
1
a
b
Figure 5.41 The set R\ (shaded area) consists of
the subrectangles of the partition of R that meet S,
the set of discontinuities of / on R. As the partition
becomes finer, the area of Ri tends toward zero.
Therefore, as all Ax, and Ay,- tend to zero, we have that the sums
J2F(xf)Axi and /<e»/).Yv,.\.\>
i RijCRi
and the term M ■ area of R\ converge (respectively) to
f F(x)dx, f f fdA, and 0.
J a J JR
332 Chapter 5 | Multiple Integration
We conclude that
that is,
5.2 Exercises
f F(x)dx - f f fdA = 0,
J a J J R
( ( fdA = f f f(x,y)dydx.
J J R J a Jc
Again, by exchanging the roles of x and y, we can show that
"d pb
as well.
[ ( fdA = f f f(x,y)dxdy
J JR Jc J a
1. Use Definition 2.3 and Theorem 2.4 to determine
the value of f fR(y3 + s'm2y)dA, where R = [0, 3] x
[-1,1]-
2. Let R = [-3, 3] x [-2, 2]. Without explicitly eval-
uating any iterated integrals, determine the value of
JfR(x5+2y)dA.
3. This problem concerns the double integral ffpX^dA,
where D is the region pictured in Figure 5.42.
(a) Determine ff£)xidA directly by setting up and ex-
plicitly evaluating an appropriate iterated integral.
(b) Now argue what the value of ffD x3 dA must be
by inspection, that is, without resorting to explicit
calculation.
Figure 5.42 The region D of
Exercise 3 .
In Exercises 4-13, evaluate the given iterated integrals. In ad-
dition, sketch the regions D that are determined by the limits
of integration.
, f f 3d
Jo Jo
r2 r1
i. I I y dx dy
Jo Jo
Jo Jo
lo Jo
^3 ?2x+\
y dy dx
-1 xydy
dx
2 + y2)dy dx
10.
11.
12.
13.
/ / {x
Jo Jx2/4
,4 ,2,/>
/ / x sin iy ) dx dy
Jo Jo
pit psinx
11 y cos x dy dx
Jo Jo
f f ' 3dydx
Jo J-JT^x1
Li zdxd>
lo J-t
y dy dx
14. Figure 5.43 shows the level curves indicating the vary-
ing depth (in feet) of a 25 ft by 50 ft swimming pool.
Use a Riemann sum to estimate, to the nearest 100 ft3,
the volume of water that the pool contains.
Figure 5.43
1 5. Integrate the function f(x, y) = 1 — xy over the trian-
gular region whose vertices are (0, 0), (2, 0), (0, 2).
5.2 I Exercises 333
16. Integrate the function f(x, y) = 3xy over the region
bounded by y = 32x3 and y = ~Jx.
17. Integrate the function f(x, y) = x + y over the region
bounded by x + y = 2 and y2 — 2y — x = 0.
18. Evaluate ffDxydA, where D is the region bounded
by x = y3 and y = x2.
19. Evaluate ffD e*2 d A, where D is the triangular region
with vertices (0, 0), (1, 0), and (1, 1).
20. Evaluate ffD 3yd A, where D is the region bounded
by xy2 = 1, y = x, x = 0, and y = 3.
21. Evaluate ffD(x — 2y)dA, where D is the region
bounded by y = x2 + 2 and y = 2x2 — 2.
22. Evaluate JfD(x2 + y2)dA, where D is the region in the
first quadrant bounded by y = x, y = 3x, and xy = 3.
23. Prove property 2 of Proposition 2.7.
24. Prove property 3 of Proposition 2.7.
25. Prove property 4 of Proposition 2.7.
26. (a) Let D be an elementary region in R2. Use the
definition of the double integral to explain why
ffD IdA gives the area of D.
(b) Use part (a) to show that the area inside a circle of
radius a isjia2.
27. Use double integrals to find the area of the region
bounded by y = x2 and y = x3.
28. Use double integrals to calculate the area of the region
bounded by y = 2x, x = 0, and y = 1 — 2x — x2.
29. Use double integrals to calculate the area inside
the ellipse whose semiaxes have lengths a and b.
(See Figure 5.44.)
Figure 5.44 The ellipse of
Exercise 29.
30. (a) Set up an appropriate iterated integral to find
the area of the region bounded by the graphs of
y = — x and y = ax2 for x > 0. (Take a to be
a constant.)
(b) Use a computer algebra system to estimate for what
value of a this area equals 1 .
31 . Use double integrals to find the total area of the region
bounded by y = x3 and x
y5-
32. Use double integrals to find the area of the region
bounded by the parabola y = 2 — x2, and the lines
X - y = 0, 2x + y = 0.
33. Let D be the region in R2 bounded by the lines x = 0,
x + y = 3, and x — y = 3. Without resorting to any
explicit calculation of an iterated integral, determine,
with explanation, the value of ffD(y3 — ex siny +
2)dA. (Hint: Use symmetry and geometry.)
34. Let D be the region in R2 with y > 0 that is
bounded by x2 + y2 = 9 and the line y = 0. Without
resorting to any explicit calculation of an iterated
integral, determine, with explanation, the value of
ffDV*
y sin* — 2) dA.
35. Determine the volume of the solid lying under the plane
z = 24 — 2x — 6y and over the region in the jcy-plane
bounded by y = 4 — x2, y = 4x — x2, and the y-axis.
36. Find the volume under the portion of the paraboloid
z = x2 + 6y2 lying over the region in the jcy-plane
bounded by y = x and y = x2 — x.
37. Find the volume under the plane z = 4x + 2y + 25
and over the region in the xy-plane bounded by y =
x2 - 10 and y = 31 - (x - \ f.
38. (a) Set up an iterated integral to compute the volume
under the hyperbolic paraboloid z = x2 — y2 + 5
and over the disk
D = {(x, y)
x2 + y2 < 4}
in the xy -plane.
(b) Use a computer algebra system to evaluate the
integral.
39. Find the volume of the region under the graph of
f(x,y) = 2-\x\-\y\
and above the xy-plane.
40. (a) Show that if R = [a, b] x [c, d], f is continuous
on [a, b], and g is continuous on [c, d], then
f(x)g(y)dA
{f,md') if.™")
(b) What can you say about
f{x)g(y)dA
11
if D is not a rectangle? More specifically, what if
D is an elementary region of type 1 ?
334 Chapter 5 Multiple Integration
41. Let
/(*. y)
1 if x is rational
0 if x is irrational and y < 1
2 if x is irrational and y > 1
(a) Show that fQ f(x,y)dy does not depend on
whether x is rational or irrational.
(b) Show that
value.
fo f(x> y)dy dx exists and find its
(c) Partition R = [0, 1] x [0, 2] and construct a
Riemann sum by choosing "test points" Cy in each
subrectangle of the partition to have rational x-
coordinates. Then to what value must this Riemann
sum converge as both Ax; and Ay j tend to zero?
( d) Partition R and construct a Riemann sum by choos-
ing test points Cy = (xf, y *) such that x* is rational
if y] < 1 and xf is irrational if y * > 1 . What hap-
pens to this Riemann sum as both A.t, and At-
tend to zero?
(e) Show that / fails to be integrable on R by us-
ing Definition 2.3. Thus, we see that double inte-
grals and iterated integrals are actually different
notions.
5.3 Changing the Order of Integration
Frequently, it is useful to think about the evaluation of double integrals over
elementary regions essentially as the determination of an appropriate order of
integration. When the region of integration is a rectangle, Fubini's theorem
(Theorem 2.6) says the order in which we integrate has no significance; that is,
fffdA=f f f(x,y)dydx=f f f(x,y)dxdy.
J J R J a J c J c J a
(See Figure 5.45.) When the region is elementary of type 1 only, we must integrate
first with respect to y (and then with respect to x) if we wish to evaluate the double
integral by means of a single iterated integral. (See Figure 5.46.) Then
fdA= / f(x,y)dydx
J JD J a Jy(x)
In the same way, when the region is elementary of type 2 only, we would typically
integrate first with respect to x, so that
r r rd rP(y)
fdA= / f(x,y)dxdy.
J JD Jc Ja(y)
®
®
®
®
R = [a, b] x [c, d]
x = a
y = 8(x)
®
®
x = b
y=y(x)
Figure 5.45 Changing the order of integration over a
rectangle.
Figure 5.46 A type 1 region
leads us to integrate with respect
to y first.
5.3 | Changing the Order of Integration 335
Figure 5.47 A type 2 region
leads to integration with respect
to x first.
Figure 5.48 The region D of
Example 1.
Figure 5.49 Integrating over
the region D of Example 1 by
integrating first with respect to x.
(See Figure 5.47.) When the region is elementary of type 3, however, we can
choose either order of integration, at least in principle. Often, this flexibility can
be used to advantage, as the following examples illustrate:
EXAMPLE 1 We calculate the area of the region shown in Figure 5.48. Con-
sidering D as a type 1 region, we obtain
Area of
p p p e p m X
D= Id A (Why?) = / / 1 dy dx
J J D J\ JO
y | Q x dx = J In x dx .
The single definite integral that results gives the area under the graph of y = In x
over the ^-interval [1, e], just as it should. To evaluate this integral, we need to
use integration by parts: Let u = In x (so du = 1 /x dx) and dv = dx (so v = x).
Then
Area of D
[
In x dx = In x ■ x\
(VI
dx
(remember f u dv = u ■ v — f v du), so
Area of D = e — 0
j dx = e — (e — 1)
1.
Integration by parts can be avoided if we integrate first with respect to x, as
schematically suggested by Figure 5.49. Hence,
Area of D
I I \dA = I I \dxdy= I x\eydy= I (e — ey)dy
J Jd Jo Je? Jo e Jo
= (ey-e?)\l0 = (e-e)-(0-e°)=l,
which checks (just as it should). ♦
Note that the two iterated integrals we used to calculate the area in Example 1 ,
namely,
/» e p In x f^fe
II dy dx and / / dxdy,
J i Jo Jo Jey
are not obtained from each other by a simple exchange of the limits of integration.
The only time such an exchange is justified is when the region of integration
is a rectangle of the form [a, b] x [c, d] so that all limits of integration are
constants.
EXAMPLE 2 Sometimes changing the order of integration can make an im-
possible calculation possible. Consider the evaluation of the following iterated
integral:
LI
y cos(x2)dx dy.
After some effort (and maybe some scratchwork), you should find it impossible
even to begin this calculation. In fact it can be shown that cos(x2) does not
have an antiderivative that can be expressed in terms of elementary functions.
Consequently, we appear to be stuck.
Chapter 5 | Multiple Integration
Figure 5.50 Note that x
corresponds to y
region shown.
fx over the
On the other hand, it is easy to integrate y cos(x2) with respect to y. This
suggests finding a way to change the order of integration. We do so in two steps:
1. Use the limits of integration in the original iterated integral to identify the
region D in R2 over which the integration takes place. (While doing this, you
should make a wish that D turns out to be a type 3 region.)
2. Assuming that the region D in Step 1 is of type 3, change the order of
integration.
The limits of integration in the preceding example imply that D can be described as
D = {(x, y) | v2 < x < 4, 0 < y < 2},
as suggested by Figure 5.50. Now Figure 5.50 can be used to change the order
of integration. We have
a:
ycos(x )dxdy
ff
Jo Jo
y cos(x )dy dx.
It is now possible to complete the calculation; that is,
ff
Jo Jo
y cos(x2) dy dx
-f
-jfi
— cos(x2)
cos(x ) dx
dx
y=0
16
cos u du,
where u = x and du = 2x dx, so that, finally,
-2
if
ycos(x )dxdy
1 • 1 16
4smMlo
sin 16.
The technique of changing the order of integration is a very powerful one, but
it is by no means a panacea for all cumbersome (or impossible) integrals. It relies
on an appropriate interaction between the integrals and the region of integration
that often fails to occur in practice.
5.3 Exercises
1 . Consider the integral
r2x
and check that your answer agrees with part (a).
In Exercises 2-9, sketch the region of integration, reverse the
order of integration, and evaluate both iterated integrals.
'-ff
Jo Jo
(2 — x — y ) dy dx
j j {2x+\)dydx. r2 r*-2*
Jo Jo
y dy dx
(a) Evaluate this integral. „2 m-/
(b) Sketch the region of integration. I I x dx dy
(c) Write an equivalent iterated integral with the order
of integration reversed. Evaluate this new integral 5. / / (x + y)dxdy
and check that your answer agrees with part (a). Jo Jjy
i. j j Idydx
5.4 | Triple Integrals
7. J J exdxdy
p7t/2 fCOSX
i. I I sinx dydx
Jo Jo
' a:
/4-y2
y dx dy
/4-v2
When you reverse the order of integration in Exercises 10 and
11, you should obtain a sum of iterated integrals. Make the
reversals and evaluate.
10. f [ (x-y)dvdx
J -2 Jx2-2
4 i-Ay-y1
11
u
(y + \)dxdy
In Exercises 12 and 13, rewrite the given sum of iterated in-
tegrals as a single iterated integral by reversing the order of
integration, and evaluate.
12
r- 1 nx p2 p2—x
. I /sin x dy dx + I I sin x dy dx
Jo Jo Ji Jo
13.
/*/
Jo Jo
8 rJyft yl2 i-^/y/i
y dx dy + I I y dx dy
Js "'Vy=S
In Exercises 14—18, evaluate the given iterated integral.
•/'/
Jo Jiy
lo ly
■a
cos (x ) dx dy
x sin xy dx dy
smx
dx dy
■ IT'
Jo Jo
■ ff •
JO Jy/2
dy dx
dx dy
It is interesting to see what a computer algebra system does with
iterated integrals that are difficult or impossible to integrate in
the order given. In Exercises 19-21, experiment with a com-
puter to evaluate the given integrals.
19. (a) Determine the value of Jfl2 J^2 y2 cos (xy) dy dx
via computer. Note how long the computer takes
to deliver the answer. Does the computer give you
a useful answer?
(b) If you were to calculate the iterated integral in part
(a) by hand, in the order it is written, what method
of integration would you use? (Don't actually carry
out the evaluation, just think about how you would
accomplish it.)
(c) Now reverse the order of integration and let your
computer evaluate this iterated integral. Does your
computer supply the answer more quickly than in
part (a)?
20. (a) See if your computer can calculate
Jo fx2 x sm (y2) dy dx as it is written,
(b) Now reverse the order of integration and have
your computer evaluate your new iterated integral.
Which of the computations in parts (a) or (b) is
easier for your computer?
^^21. (a) Can your computer evaluate /J f^-i ecosxdx dy?
(b) Reverse the order of integration and have it try
again. What happens?
f— y
Figure 5.51 The box
B = [a, b] x [c, d] x [p, q].
5.4 Triple Integrals
Let f{x, y, z) be a function of three variables. Analogous to the double integral,
we define the triple integral of / over a solid region in space to be the limit of
appropriate Riemann sums. We begin by denning this integral over box-shaped
regions and then proceed to define the integral over more general solid regions.
The Integral over a Box
Let B be a closed box in R3 whose faces are parallel to the coordinate planes.
That is,
B = {{x, y, z) eR3 | a < x < b, c < y <d, p < z <q}.
(See Figure 5.51.) We also use the following shorthand notation for B:
B = [a, b] x [c, d] x [p, q].
Chapter 5 | Multiple Integration
b=x„
c = y0 y\ yi
y
q = Zn
P = Zo
d = yn
Figure 5.52 A partitioned box.
DEFINITION 4.1 A partition of B of order n consists of three collections
of partition points that break up B into a union of n3 subboxes. That is, for
i, j, k = 0, . . . , n, we introduce the collections {jc, }, {yj}, and {zk}, such that
a = xo < x\ < ■ ■ ■ < < Xj < ■ • • < x„ = b,
c = yo < y\ < ■ ■ ■ < yj-i < yj < ■■■ < yn = d,
p = Zo < Z\ < ■ ■ ■ < Zk-l < Zk < ■ ■ ■ < z„ = q.
(See Figure 5.52.) In addition, for i, j,k = 1 n, let
Axj=Xi—Xi-u Ayj = yj — yj-\, and Azk = Zk - Zk-u
DEFINITION 4.2 Let / be any function defined on B = [a, b] x [c, d] x
[p, q]. Partition B in some way. Let Cy* be any point in the subbox
Bijk = [xt-i,Xi] x [yj-uyj] x [z*-i. Zifc] (},j,k= I,..., n).
Then the quantity
n
S= f(cljk)AVijk,
i,j,k=l
where AVfjk = AxjAyjAzk is the volume of B^, is called the Riemann
sum of j f on B corresponding to the partition.
You can think of the Riemann sum ^ f(Cijk)AVijk as a weighted sum of
volumes of subboxes of B, the weighting given by the value of the function / at
particular "test points" Cy£ in each subbox.
DEFINITION 4.3 The triple integral of / on £, denoted by
///.'"•
by jjj f(x,y,z)dV, orby jjj f(x,y,z)dxdydz,
5.4 | Triple Integrals
/
o
/
/
/
B
Figure 5.53 The subbox
contributes /(c,/a-)A V;,^
to the Riemann sum S. If
we think of / as
representing a density
function, then the total
mass of the entire box B
KffhfdV.
is the limit of the Riemann sum S as the dimensions Ax, , Ay, , and Azk of
the subboxes Bjjk all approach zero, that is,
fff fdV= lim T /(c,;A)A v, Ay, A-A.
J J JB allA«,A^,A«-.0,j^1
provided that this limit exists. When fffBfdV exists, we say that / is
integrable on B.
The key point to remember is that the triple integral is the limit of Riemann
sums. It is this notion that enables useful and important applications of integrals.
For example, if we view the integrand / as a type of generalized density function
("generalized" because we allow negative density!), then the Riemann sum S is
a sum of approximate masses (densities times volumes) of subboxes of B. These
approximations should improve as the subboxes become smaller and smaller.
Hence, we can use the triple integral fffg f dV, when it exists, to compute the
total mass of a solid box B whose density varies according to /, as suggested by
Figure 5.53.
Analogous to Theorem 2.5, we have the following result regarding integra-
bility of functions:
THEOREM 4.4 If / is bounded on B and the set of discontinuities of / on B
has zero volume, then fffBfdV exists. (See Figure 5.54.)
Figure 5.54 In
Theorem 4.4 the
discontinuities of / on B
(shown shaded) must have
zero volume.
To say that a set X has zero volume as we do in Theorem 4.4, we mean that
we can cover X with boxes B\, B2 Bn, . . . (i.e., so that X c [J™=1 Bn), the
sum of whose volumes can be made arbitrarily small.
To evaluate a triple integral over a box, we can use a three-dimensional version
of Fubini's theorem.
THEOREM 4.5 (Fubini's theorem) Let / be bounded on B = [a, b] x
[c, d] x [p, q] and assume that the set S of discontinuities of / has zero vol-
ume. If every line parallel to the coordinate axes meets S in at most finitely many
points, then
///.
fdV
pi? pel pq pb pq pel
— f{x,y,z)dzdydx= III
J a J c J p J a J p J c
pd pb pq pd I pq pb
= / / / f(x,y,z)dzdxdy= / / / f(x,y,z)dxdzdy
J c J a J p J c J p J a
f(x, y, z)dy dz dx
pq pb pd pq pd pb
= I I I f(x,y,z)dydxdz= III f(x,y,z)dxdydz.
J p J a J c J p J c J a
q pd pb
340 Chapter 5 | Multiple Integration
Figure 5.55 The function / is
continuous on W.
z=Y(x,y)
z = (p(x,y)
-y
Figure 5.56 An elementary
region of type 1 .
y = d
x = a(y){^
y = c
Figure 5.57
The "shadow"
(projection) of W into the .rj-plane
should be an elementary region in
the plane.
EXAMPLE 1 Let
B = [-2, 3] x [0, 1] x [0, 5], and let f(x, y, z) = x2ey + xyz.
Thus, / is continuous and hence certainly satisfies the hypotheses of Fubini's
theorem. Therefore,
f f [ (xV +xyz)dV = f f f (x2ey +xyz)dzdydx
J J Jb J -2 Jo JO
J (x2eyz + \xyz2)\l=0 dy dx
= j j (5x2ey + f xy) dy dx
= ^ {5x2ey + 2ixy%T^dx
= j (5(e- l)x2 + fx)dx
= (f(e-iy + f^)|3_2
= (45(e-l)+f)-(-f(e-l)+f)
= f(,-D+f.
You can check that integrating in any of the other five possible orders produces
the same result. ♦
Elementary Regions in Space
Now suppose W denotes a fairly arbitrary solid region in space, like a rock or a
slab of tofu. Suppose / is a continuous function defined on W, such as a mass
density function. (See Figure 5.55.) Then the triple integral of / over W should
give the total mass of W. As was the case with general double integrals, we need to
find a way to properly define fffw f dV and to calculate it in practical situations.
The course of action is much like before: We see how to calculate integrals over
certain types of elementary regions and treat integrals over more general regions
by subdividing them into regions of elementary type.
DEFINITION 4.6 We say that W is an elementary region in space if it can
be described as a subset of R3 of one of the following four types:
Type 1 (see Figures 5.56 and 5.57)
(a) W = {(x, y, z) | <p(x, y) < z < f(x, y), y(x) < y < S(x), a <x <b],
or
(b) W = {(x, y, z) | <p(x, y)<z< fix, y), a(y) <x< B(y), c<y<d}.
5.4 | Triple Integrals 341
Type 2 (see Figure 5.58)
(a) W = {(x, y, z) | a(y, z) < x < P(y, z), y(z) <y< S(z), p<z<q),
or
(b) W = {(x, y, z) | a(y, z) < x < p(y, z), <p(y) <z< f(y), c<y< d}.
Type 3 (see Figure 5.59)
(a) W = {(x, y, z) | y(x, z) <y < S(x, z), a(z) <x< fi(z), p < z < q},
or
(b) W = {(x, y, z) | y{x, z)< y < S(x, z), (p(x) <z< ir(x), a<x<b}.
Type 4
W is of all three previously described types.
x= a(y,z)
y=y(x,z)
y = 5(x, z)
Figure 5.58 For an elementary
region of type 2, the shadow in the
jz-plane should be an elementary
region in the plane.
Figure 5.59 For an elementary
region of type 3, the shadow in the
-tz-plane should be an elementary
region in the plane.
Some explanation regarding Definition 4.6 is in order. An elementary region
W of type 1 is a solid shape whose top and bottom boundary surfaces each can
be described with equations that give z as functions of x and y and such that
the projection of W into the xy-plane (the "shadow") is in turn an elementary
region in R2 (in the sense of Definition 2.8). Similarly, an elementary region of
type 2 is one whose front and back boundary surfaces each can be described with
equations giving x as functions of y and z and whose projection into the yz-plane
is an elementary region in R2 . Finally, an elementary region of type 3 is one whose
left and right boundary surfaces each can be described with equations giving y
as functions of x and z and whose projection into the xz-plane is an elementary
region in R2. In each case, an elementary region in space is one for which we
can find boundary surfaces described by equations where one of the variables is
expressed in terms of the other two, and whose "shadow" in the plane of these
two variables is an elementary region in R2 in the sense of Definition 2.8.
EXAMPLE 2 Let W be the solid region bounded by the hemisphere x2 + y2 +
z2 = 4, where z < 0, and the paraboloid z = 4 — x2 — y2. (See Figure 5.60.) It
is an elementary region of type 1 since we may describe it as
W = {(x, y, z) | -x2-y2 < z < 4
-V4 - x2 < y < ^4 - x2, -2 < x < 2j
2 2
x - y ,
342 Chapter 5 | Multiple Integration
z
x2 + y2 + z2 = 4,
z < 0 Shadow of W
Figure 5.60 The solid region W of Example 2.
This description was obtained by noting that W is bounded on top and bottom by a
pair of surfaces, each of which is the graph of a function of the form z = g(x, y)
and the shadow of W in the xy-plane is a disk D of radius 2, which we have
chosen to describe as
D=\(x,y) | -J*-
< y <
-2 < x < 2
and which we already know is an elementary region (of type 3) in the xy-
plane. ♦
EXAMPLE 3 The solid bounded by the ellipsoid
a,b,c positive constants
2 2 2
^ XL V Z
E : — + J— + —
b2
1.
can be seen to be an elementary region of type 4. To see that it is of type 1 , split
the boundary surface in half via the z = 0 plane as shown in Figure 5.61. (This
is accomplished analytically by solving for z in the equation for the ellipsoid.)
Then the shadow D of E is the region inside the ellipse in the xy-plane shown in
Figure 5.62.
Figure 5.61 The ellipsoid of
Example 3 as an elementary
region of type 1 .
Figure 5.62 The shadow of
the type 1 ellipsoid in
Figure 5.61 is the region inside
the ellipse x2/a2 + y2/b2 = 1
in the xy-plane.
5.4 | Triple Integrals 343
Figure 5.63 The ellipsoid of
Example 3 as a type 2 elementary
region.
Figure 5.64 The shadow
of the ellipsoid in
Figure 5.63 is the region
inside the ellipse
y2/b2 + z2/c2 = 1 in the
yz-plane.
We have
D = \(x, y)
Figure 5.65 The ellipsoid of
Example 3 as a type 3 region.
-bJ\
y < bJi
-a < x < a
= (*, y)
-a
'l-^, -b< y< b\
Figure 5.66 The
shadow of the
ellipsoid in
Figure 5.65 in the
Az-plane.
so D is in fact a type 3 elementary region in R2.
To see that E is of type 2, split the boundary at the x = 0 plane as in
Figure 5.63. The shadow in the yz-plane is again the region inside an ellipse.
(See Figure 5.64.) Finally, to see that E is of type 3, split along y = 0. (See
Figures 5.65 and 5.66.) ♦
Triple Integrals in General —
Suppose W is an elementary region in R3 and / is a continuous function on W.
Then, just as in the case of double integrals, we define the extension of / by
.f(x,y,z) if(x,y,z)eW
f (x, y,z) = {
0 if(x,y,z)£W
By Theorem 4.4, fext is integrable on any box B that contains W. Thus, we can
make the following definition:
DEFINITION 4.7 Under the assumptions that W is an elementary region
and / is continuous on W, we define the triple integral
fff fdV tobe fff rxdV,
J J Jw J J J B
where B is any box containing W.
344 Chapter 5 | Multiple Integration
Using a proof analogous to that of Theorem 2.10, we can establish the
following:
THEOREM 4.8 Let W be an elementary region in R3 and / a continuous
function on W.
1. If W is of type 1 (as described in Definition 4.6), then
err fb rs(x) rirQc>y)
/// fdV= / / f(x,y,z)dzdydx, (type la)
J J JW J a Jy(x) J(fi(x,y)
or
r r r rd rP(y) rf(x>y)
ill fdV= / / f(x,y,z)dzdxdy. (type lb)
J J JW Jc Ja(y) J<p(x,y)
2. If W is of type 2, then
r r r rl r$(z) rP(y,z)
fdV= / f(x,y,z)dxdydz,
J J JW J p Jy(z) Ja(y,z.)
or
r r r rd rf(y) rP(y,z)
// / fdV= / / f(x,y,z)dxdzdy. (type 2b)
J J JW Jc Jtp(y) Ja(y,z.)
3. If W is of type 3, then
or
r r r ri rP(z) rHx,z)
fdV= / f(x,y,z)dydxdz,
J J JW J p Ja(z) Jy(x,z)
err rb rfU) rHx,z.)
fdV= / / f(x,y,z)dydzdx.
J J JW Ja Jtp(x) Jy(x,z)
(type 2a)
(type 3a)
(type 3b)
(0,0, 1).
Plane x + y + z = 1
.(0,1,0)
-y
(1,0, 0)
Figure 5.67 The tetrahedron of
Example 4.
EXAMPLE 4 Let W denote the (solid) tetrahedron with vertices at (0, 0, 0),
(1,0, 0), (0, 1, 0), and (0, 0, 1) as shown in Figure 5.67. Suppose that the mass
density at a point (x, y, z) inside the tetrahedron varies as f(x, y, z) — 1 + xy.
We will use a triple integral to find the total mass of the tetrahedron.
The total mass M is
HI fdV= fff (l+xy)dV.
J J Jw J J Jw
(See the remark before Theorem 4.4.) To evaluate this triple integral using iterated
integrals, note that we can view the tetrahedron as a type 1 elementary region.
(Actually, it is a type 4 region, but that will not matter.) The slanted face is given
by the equation x + y + z = 1, which describes the plane that contains the three
points (1, 0, 0), (0, 1, 0), and (0, 0, 1). Hence, by first integrating with respect to
5.4 | Triple Integrals 345
z and holding x and y constant,
(0, 1, 0) ,
* Line x + y = 1
\^ (in z = 0 plane)
(1,0,0)
Figure 5.68 The shadow in the
xy-plane of the tetrahedron of
Example 4 is a triangular region.
Figure 5.69 The region W of
Example 5.
M= f f (f \\+xy)dz\dA
J J shadow \J 0 /
= 11 {l+xy){\-x-y)dA
J J shadow
= / / (l - x - y + xy - x2y - xy2) dA.
J J shadow
The shadow of W in the x v-plane is just the triangular region shown in Figure 5.68.
Thus,
M = (l - x — y + xy - x2y - xy2) dA
J J shadow
= / / (l — x — y + xy — x2y — xy2) dy dx
Jo Jo
= f ((1 - je) - x(\ - x) - 1(1 - x)2 + \x{\ - x f
Jo
10
L
lx2(l - xf - \x{\-xf)dx
= G-f* + ^3-|*V*=(2
1 _ 5^ 1 .3 _ lx4)dx = (lx _ + lx4 _ l_x5yL _ 1_
12'
30 )\0 40"
Note that M can also be written as a single iterated integral, namely,
M
pi p 1— x p 1
Jo Jo Jo
(1 + xy)dz dy dx.
EXAMPLE 5 We calculate the volume of the solid W sitting in the first octant
and bounded by the coordinate planes, the paraboloid z = x2 + y2 + 9, and the
parabolic cylinder y = 4 — x2. (See Figure 5.69.)
By definition, the triple integral is a limit of a weighted sum of volumes of
tiny subboxes that fill out the region of integration. If the weights in the sum are
all taken to be 1, then we obtain an approximation to the volume:
i,j,k
Therefore, taking the limit as the dimensions of the subboxes all approach zero,
it makes sense to define
v= ////""■
Chapter 5 Multiple Integration
(0,4)
Figure 5.70 The shadow
in the xy-plane of the
region in Figure 5.69.
z = 3x2 + 3y2-16
gure 5.71 The capsule-shaped
gion of Example 6.
Figure 5.72 The shadow
of the region W of
Example 6, obtained by
projecting the intersection
curves of the defining
paraboloids onto the
xy-plane.
In our situation, W is a type 1 region whose shadow in the xy-plane looks
like the region shown in Figure 5.70. Thus, by Theorem 4.4,
V = dV = / / dzdydx
J J Jw Jo Jo Jo
= / / (x2 + y2 + 9)dydx
Jo Jo
= f*(*ly + \ys + 9y\^)dx
= f (x2(4 - x2) + i(4 - x2)3 + 9(4 - x2)) dx
Jo
= f2 (m-2lx2 + 3x4 - \x6)dx
= (ipx _ 7x3 + fx
J_v7)l -
21 "
35
EXAMPLE 6 We find the volume inside the capsule bounded by the paraboloids
z = 9 — x2 — y2 and z = 3x2 + 3y2 — 16. (See Figure 5.71.)
Once again, we have
V
IdV,
and again the region W of interest is elementary of type 1 . The shadow, or pro-
jection, of W in the xy-plane is determined by
{(x, y) e R2 | there is some z such that (x, y, z) e W}.
Physically, one can also imagine the shadow as the hole produced by allowing W
to "fall through" the xy-plane. In other words, the shadow is the widest part of W
perpendicular to the z-axis. From Figure 5.71, one can see that it is determined
by the intersection of the two boundary paraboloids. The shadow itself is shown
in Figure 5.72. The intersection may be obtained by equating the z-coordinates
of the boundary paraboloids. Therefore,
9 - x2 - y2 = 3x2 + 3y2
Thus, by Theorem 4.4,
16
4x2 + 4y2 = 25
*2 + y2 = 2{ = (l)2-
f f f fS/2 r^/25/4-x2 p9-x2-y2
V = dV = / / dzdydx
J J Jw J -5/2 J-Jli/A-x2 ^3jc2+3y2-16
/•5/2 p^/25/4-x2 p9-x2-y2
= 4 / / / dz dy dx.
Jo Jo J3x2+3v2
v2-16
This last iterated integral represents the volume of one quarter of the capsule.
Hence, we multiply its value by 4 to obtain the total volume. The reason for this
5.4 I Exercises 347
manipulation is to make the subsequent calculations somewhat simpler (although
the computation that follows is clearly best left to a computer).
We compute
V
rS/2 *9-x*-f
= 41 / / dzdydx
J0 J0 Jix2+3y2-i6
r5/2 p^25/4-x2
= 4 / / (25 -4x2 -4y2)dydx
Jo Jo
(f-x2)3/2)jx
= 4^ U25-4x%
/•5/2 /
2^ /25 _ x2 _ 4 ^25 _ v2\3/2
25 _ r2 _ 4 ^25
4 A 3
(f -x2)3/2)jx
(f -x2)3/2)jx
= 4 (4-f)(f-x2)3^x
Jo
f5/2
= ¥/ (2i-x2f2dX.
Jo
Now let x = J sin#, so dx = | cos9 d9. Then
32 f^2 /5 \3 5 1250 f^2
32 f^2 /5 V 5 1250 Z^2 4
V = — -cosS -cos6d8 = / cos4 Ode
3 Jo \2 / 2 3 Jo
rn/2 (
L G
1250 Z^2 /l \2
— / (-(1+COS20) ^
625 /'7r/2 ,
/ (1 + 2 cos 2(9 + cos2 26) d6
6 Jo
625
~6~
625 r1* ' \
(6» + sin26»)|o/2 + — / -(1 + cos 46) d6
6 Jo 2
12 (H =
625 625 (it \ 625n
7t +
12
5.4 Exercises
Evaluate the triple integrals given in Exercises 1—3.
1. f [ [ xyzdV
J J J[-l,l]x[0,2]x[l,3]
2 SSL
SSL
(x2 + y2 + z2)dV
3. I I dV
/[l,e]x[l,e]x[l,e]
4. Find the value of fffwzdV, where W = [—1, 2] x
[2, 5] x [—3, 3], without resorting to explicit
calculation.
/[0,l]x[0,2]x[0,3]
348 Chapter 5 | Multiple Integration
Evaluate the iterated integrals given in Exercises 5—7.
5. f I [ 3yz2dxdydz
J-i Ji Jo
6. J I I (x + 2y + z) dy dx dz
J\ Jo Ji
ply ry+z
JO Jl+y Jz
z dx dz dy
8. (a) Let W be an elementary region in R3. Use the
definition of the triple integral to explain why
fffw 1 dV gives the volume of W.
(b) Use part (a) to find the volume of the region
W bounded by the surfaces z = x2 + y2 and z =
9-x2 -y2.
9. Use triple integrals to verify that the volume of a ball
of radius a is 4jra3/3.
1 0. Use triple integrals to calculate the volume of a cone of
radius r and height It . (You may wish to use a computer
algebra system for the evaluation.)
In Exercises 11—20, integrate the given function over the indi-
cated region W.
11. f(x, y, z) — 2x — y + z; W is the region bounded
by the cylinder z = y2, the Jty-plane, and the planes
x = 0, x = 1, y = —2, y = 2.
12. f(x,y,z) = y; W is the region bounded by the
plane x + y + z = 2, the cylinder x2 + z2 = 1, and
y = 0.
13. f(x,y,z) = Sxyz; W is the region bounded by
the cylinder y = x2, the plane y + z = 9, and the
xy-plane.
14. f(x,y,z) = z", W is the region in the first octant
bounded by the cylinder y2 + z2 = 9 and the planes
y = x, x = 0, and z = 0.
15. f(x, y,z) = 1 — z2', W is the tetrahedron with vertices
(0, 0, 0), (1, 0, 0), (0, 2, 0), and (0, 0, 3).
16. f(x, y, z) = 3.x:; W is the region in the first octant
bounded by z = x2 + y2, x = 0, y = 0, and z = 4.
17. f(x, y, z) = x + y; W is the region bounded by
the cylinder x2 + 3z2 = 9 and the planes y = 0,
x + y = 3.
18. f(x, y, z) = z; W is the region bounded by z = 0,
x2 + Ay2 = 4, and z = X + 2.
19. f(x, y, z) = 4jc + y; W is the region bounded by x =
y2 , y = z, x = y, and z = 0.
20. f(x, y, z) = x; W is the region in the first octant
21. Find the volume of the solid bounded by z = 4 — x2,
x + y = 2, and the coordinate planes.
22. Find the volume of the solid bounded by the planes y =
0, z = 0, 2y + z = 6, and the cylinder x2 + y2 = 9.
23. Find the volume of the solid bounded by the paraboloid
z = Ax2 + y2 and the cylinder y2 + z = 2.
24. Find the volume of the region inside both of the cylin-
ders x2 + y2 = a2 and x2 + z2 = a2.
25. Consider the iterated integral
• l-x
lo
f(x, y, z)dz dx dy.
Sketch the region of integration and rewrite the inte-
gral as an equivalent iterated integral in each of the five
other orders of integration.
26. Change the order of integration of
f(x, y, z)dz dx dy
fff
Jo Jo Jo
to give five other equivalent iterated integrals.
27. Change the order of integration of
fff
Jo Jo Jo
f(x, y, z)dz dy dx
bounded by z = x + 2y , z = 6
and y = 0.
y , x
o,
to give five other equivalent iterated integrals.
28. Consider the iterated integral
/ / / 2dzdydx.
Jo Jo Ay2
(a) This integral is equal to a triple integral over a solid
region W in R3. Describe W.
(b) Set up an equivalent iterated integral by integrating
first with respect to z, then with respect to x, then
with respect to y. Do not evaluate your answer.
(c) Set up an equivalent iterated integral by integrating
first with respect to y, then with respect to z, then
with respect to x. Do not evaluate your answer.
(d) Now consider integrating first with respect to y,
then x, then z. Set up a sum of iterated integrals
that, when evaluated, give the same result. Do not
evaluate your answer.
(e) Repeat part (d) for integration first with respect to
x, then z, then y.
29. Consider the iterated integral
,2 n\sf4^ rA-y1
/ / / (x3 +y3)dzdydx.
3-2 Jo Jx2+3y2
(a) This integral is equal to a triple integral over a solid
region W in R3. Describe W.
5.5 I Change of Variables 349
(b) Set up an equivalent iterated integral by integrating
first with respect to z, then with respect to x, then
with respect to y. Do not evaluate your answer.
(c) Set up an equivalent iterated integral by integrating
first with respect to x, then with respect to z, then
with respect to y. Do not evaluate your answer.
(d) Now consider integrating first with respect to x,
then y, then z. Set up a sum of iterated integrals
that, when evaluated, give the same result. Do not
evaluate your answer.
(e) Repeat part (d) for integration first with respect to
y, then z, then x.
5.5 Change of Variables
As some of the examples in the previous sections suggest, the evaluation of a mul-
tiple integral by means of iterated integrals can be a complicated process. Both the
integrand and the region of integration can contribute computational difficulties.
Our goal for this section is to see ways in which changes in coordinates can be
used to transform iterated integrals into ones that are relatively straightforward to
calculate. We begin by studying the coordinate transformations themselves and
how such transformations affect the relevant integrals.
Coordinate Transformations
Let T: R2 —> R2 be a map of class C1 that transforms the wu-plane into the xy-
plane. We are interested particularly in how certain subsets D* of the wu-plane
are distorted under T into subsets D of the xy-plane. (See Figure 5.73.)
D*
CO
D = T(D*)
Figure 5.73 The transformation T(m, v) = (x(u, v), y(u, v))
takes the subset D* in the uv-p\am to the subset D = {(x, y) |
(x, y) = T(m, v) for some (u, ») e D*) of the xy-plane.
EXAMPLE 1 Let T(u, v) = (u + 1 , v + 2); that is, let x = u + 1, y = v + 2.
This transformation translates the origin in the Mu-plane to the point (1 , 2) in the
xy-plane and shifts all other points accordingly. The unit square D* = [0, 1] x
[0, 1 ], for example, is shifted one unit to the right and two units up but is otherwise
unchanged as shown in Figure 5.74. Thus, the image of D* is D = [1,2] x [2,3].
♦
EXAMPLE 2 Let S(w, v) = (2m, 3d). The origin is left fixed, but S stretches all
other points by a factor of two in the horizontal direction and by a factor of three
in the vertical direction. (See Figure 5.75.) ♦
EXAMPLE 3 Composing the transformations in Examples 1 and 2, we obtain
(T o S)(«, «) = T(2«, 3u) = (2m + 1, 3v + 2).
Such a transformation must both stretch and translate as shown in Figure 5.76. ♦
350 Chapter 5 | Multiple Integration
D
D
□
= T(D')
Figure 5.74 The image of D* = [0, 1] x [0, 1] is
D = [1, 2] x [2, 3] under the translation
T(m, v) = (u + 1, v + 2) of Example 1.
£>* = [0, 1] x [0, 1]
D = S(D*) = [0, 2] x [0, 3]
Figure 5.75 The transformation S of Example 2 is a scaling
by a factor of 2 in the horizontal direction and 3 in the vertical
direction.
y
ToS
D
D = [1, 3] x [2, 5]
1
Figure 5.76 Composition of the transformations of
Examples 1 and 2.
EXAMPLE 4 Let T(w, v) = (u + v, u — v). Because each of the component
functions of T involves both variables u and v, it is less obvious how the unit
square D* = [0, 1] x [0, 1] transforms. We can begin to get some idea of the
geometry by seeing how T maps the edges of D* :
Bottom edge: (u, 0), 0 < u < 1
Top edge: (u, 1), 0 < u < 1
Left edge: (0, v), 0 < v < 1
Right edge: (1, v), 0 < v < 1
T(m, 0) = (u, u);
T(m, 1) = (u + 1, m — 1);
T(0, v) = (v, -v);
T(l, w) = O + l, 1 - v).
By sketching the images of the edges, it is now plausible that the image of D*
under T is as shown in Figure 5.77. ♦
Figure 5.77 The transformation T of Example 4.
by
5.5 | Change of Variables 351
More generally, we consider linear transformations T: R2 — > R2 defined
T(w, v) = (au + bv, cu + dv)
a b
u
c d
V
where a, b, c, and d are constants. (Note: The vector (u, v) is identified with the
u
v
2 x 1 matrix
.) One general result is stated in the following proposition:
PROPOSITION 5.1 Let A =
fined by
a b
c d
, where det A ^ 0. If T: R2 -> R2 is de-
T(w, w) = A
then T is one-one, onto, takes parallelograms to parallelograms and the vertices
of parallelograms to vertices. (See §2. 1 to review the notions of one-one and onto
functions.) Moreover, if D* is a parallelogram in the i/u-plane that is mapped
onto the parallelogram D = T(D*) in the xy-plane, then
Area of D = \ det A| ■ (Area of D*).
EXAMPLE 5 We may write the transformation T(u, v) = (u + v, u
Example 4 as
v) in
T(m, v)
Note that
det
u
v
= -2^0.
Hence, Proposition 5.1 tells us that the square D* = [0, 1] x [0, 1] must be
mapped to a parallelogram D = T(Z)*) whose vertices are
T(0, 0) = (0, 0), T(0, 1) = (1, -1), T(1,0) = (1,1), T(l, 1) = (2,0).
Therefore, Figure 5 .77 is indeed correct and, in view of Proposition 5.1, could have
been arrived at quite quickly. Also note that the area of D is | — 2| • 1 = 2. ♦
Proof of Proposition 5.1 First we show that T is one-one. So suppose
T(w, v) = T(u', v'). We show that then u = u',v = v' . We have
T(w, v) = T(u' , v')
if and only if
(au + bu , cu + dv) = (au' + bv', cu' + du').
By equating components and manipulating, we see this is equivalent to the system
a(u — u') + b(v — v') = 0
c(u - u') + d(v - v') = 0
(1)
352 Chapter 5 | Multiple Integration
If a 7^ 0, then we may use the first equation to solve for u — u':
u! = — (v- v')
(2)
Figure 5.78 The vertices of
D* = {p + sn + tb | 0 < s,t <
1 } are at p, p + a, p + b,
p + a + b (i.e., where s and t take
on the values 0 or 1).
Figure 5.79 The image D of the
parallelogram D* under the linear
transformation T(u) = An.
Hence, the second equation in (1) becomes
be
(v - v') + d(v -v') = 0
or, equivalently,
-be + ad
(v - v') = 0.
By hypothesis, det A = ad — be ^ 0. Thus, we must have v — v' = 0 and there-
fore, u — u' = 0 by equation (2). If a = 0, then we must have both b 0 and
c 0, since det A ^ 0. Consequently, the system (1) becomes
J b(v -v') =0
I c(u - u') + d(v - v') = 0 "
The first equation implies v — v' = 0 and hence, the second becomes c(u — u') =
0, which in turn implies u — u! = 0, as desired.
To see that T is onto, we must show that, given any point (x, y) e R2, we can
find (u, v) e R2 such that T(m, v) = (x, y). This is equivalent to solving the pair
of equations
Iau + bv = x
cu + dv = y
for u and v. We leave it to you to check that
dx — by ay — cx
14 = ~~a — TT and v = ~1 — TT
ad — be ad — be
will work.
Now, let D* be a parallelogram in the ww-plane. (See Figure 5.78.) Then D*
may be described as
D* = {u | u = p + + tb, 0 < s < 1, 0 < t < 1}.
Hence,
D = T(D*) = {Au ueD)
= {A(p + 5a + rb) | 0 < s < 1, 0<?<1}
= {Ap + ^Aa + rAb | 0 < s < 1, 0 < t < 1}.
If we let p' = Ap, a' = Aa, and b' = Ab, then
D = {p' + sa' + rb' | 0 < s < 1,0 < r < 1}.
Thus, £> is also a parallelogram and moreover, the vertices of D correspond to
those of D*. (See Figure 5.79.)
Finally, note that the area of the parallelogram D* whose sides are parallel to
a2
and b
by
b2
5.5 | Change of Variables
may be computed as follows:
j
k
Areaof/?* = ||axb|| =
det
0
bi
b2
0
= \a.\b2 — a2b\
Similarly, the area of D = T(D*) whose sides are parallel to
a =
and b' =
b\
b'
is
a'2b\ j
a
b
a\
aa\ -
- ba2
. a2 .
c
d
CL2
ca\ -
\-da2
' b\ '
a
b ~
" bx '
ab\ -
Ybb2
,b2_
c
d
. b2 .
cb\ -
Ydb2
Area of D = ||a' x b'|| = \a[b'2
Now, a' = Aa and b' = Ab. Therefore,
and
Hence, by appropriate substitution and algebra,
Area of D = \(aai + ba2)(cb\ + db2) — (ca\ + da-£)(ab\ + bb2)\
= \{ad — bc)(a\b2 — a2b\)\
= I det A] • area of D*.
Note that we have not precluded the possibility of D*'s being a "degenerate"
parallelogram, that is, such that the adjacent sides are represented by vectors a
and b, where b is a scalar multiple of a. When this happens, D will also be
a degenerate parallelogram. The assumption that det A 7^ 0 guarantees that a
nondegenerate parallelogram D* will be transformed into another nondegenerate
parallelogram, although we have not proved this fact. ■
Essentially all of the preceding comments can be adapted to the three-
dimensional case. We omit the formalism and, instead, briefly discuss an example.
EXAMPLE 6 Let T: R3 -> R3 be given by
T(m, v, w) = (2u, 2u + 3v + w, 3w).
Then we rewrite T by using matrix multiplication:
Note that if
- 2
0
0 "
u
w)
2
3
1
V
_ 0
0
3 _
w
" 2
0
0 "
\ =
2
3
1
»
0
0
3
then det A = 18^0.
354 Chapter 5 | Multiple Integration
A result analogous to Proposition 5.1 allows us to conclude that T is one-one
and onto, and T maps parallelepipeds to parallelepipeds. In particular, the unit
cube
D* = [0, 1] x [0, 1] x [0, 1]
is mapped onto some parallelepiped D = T(£>*) and, moreover, the volume of D
must be
| det A| - volume of D* = 18- 1 = 18.
To determine D, we need only determine the images of the vertices of the cube:
r(o, o, 0) = (o, o, o)
T(0, 0, 1) = (0, 1,3)
T(0, 1, 1) = (0,4, 3)
Both D* and its image D are shown in Figure 5.80.
7X1, 0, 0) = (2, 2, 0); T(0, 1,0) = (0, 3, 0);
T(l, 1, 0) = (2, 5, 0); 7X1, 0, 1) = (2, 3, 3);
J(l, 1, 1) = (2, 6, 3).
D*
Figure 5.80 The cube D* and its image D under the
linear transformation of Example 6.
EXAMPLE 7 Of course, not all transformations are linear ones. Consider
(x, y) = T(r, 9) = (r cos#, r sin#).
Note that T is not one-one since T(0, 0) = (0, 0) = T(0, n). (Indeed T(0, 9) =
(0, 0) for all real numbers 9.) Note that vertical lines in the r0-plane given by
r = a, where a is constant, are mapped to the points (x, y) = (a cos#, a sin#)
on a circle of radius a. Horizontal rays {(r, 9) \ 9 = a, r > 0} are mapped to
rays emanating from the origin. (See Figure 5.81.) It follows that the rectangle
D* = [5, 1] x [0, it] in the r#-plane is mapped not to a parallelogram, but bent
6
G = a
r = a
Image of
6 = a
Image of
r = a
Figure 5.81 The images of lines in the r#-plane under the
transformation T(r, 6) = (r cos 0, r sin 0).
5.5 | Change of Variables 355
D
Figure 5.82 The image of the rectangle D* = [±, 1] x [0, n]
under T(r, 9) = (r cos 6, r sin#).
Figure 5.83 The image of B* = [\, 1] x [0, it] x [0, 1]
under T(r, 6, z) = (r cos 6, r sin#, z).
into a region D that is part of the annular region between circles of radii \ and 1 ,
as shown in Figure 5.82.
Analogously, the transformation T: R3 — > R3 given by
(x, y, z) = T(r, 6, z) = (r cos#, r sin^, z)
bends the solid box B* = [j, 1] x [0, 7r] x [0, 1] into a horseshoe-shaped solid.
(See Figure 5.83.) ♦
Change of Variables in Definite Integrals
Now we see what effect a coordinate transformation can have on integrals and how
to take advantage of such an effect. To begin, consider a case with which you are
already familiar, namely, the method of substitution in single-variable integrals.
EXAMPLE 8 Consider the definite integral /Q2 2x cos(x2) dx. To evaluate, one
typically makes the substitution u = x2 (so du = 2x dx). Doing so, we have
sin 4.
/ 2x cos(x2) dx = I cosudu = sinu
Jo Jo
u=4
u=0
Let's dissect this example more carefully. First of all, the substitution u = x2
may be rewritten (restricting x to nonnegative values only) as x = *Ju. Then
dx = du /(ly/u) and
f2 f4 du f4
/ 2xcos(x2)dx= I 2yfu cos(a/m)2 — — = / cosMJw = sin4.
Jo Jo 2y/u Jo
In other words, the substitution is such that the 2x = 2^fu factor in the integrand
is canceled by the functional part of the differential dx = du/(2*Ju). Hence, a
simple integral results. ♦
In general, the method of substitution works as follows: Given a (perhaps
complicated) definite integral fA f(x)dx, make the substitutions = x(u), where
x is of class C1. Thus, dx = x'(u)du. If A = x(a), B = x(b), and x'(u) 7^ 0 for
u between a and b, then
/• B nb
I f(x)dx = J f(x(u))x'(u)du. (3)
J A Ja
Chapter 5 | Multiple Integration
x
x = x(u)
It
Figure 5.84 As Am = du — >• 0,
Ax — > dx = x'(u) Am. Thus, the factor
x'(u) measures how length in the w-direction
relates to length in the x-direction.
Note that it is possible to have a > b in (3) above. (This happens if x(u) is
decreasing.) Although the m -integral in equation (3) may at first appear to be more
complicated than the x -integral, Example 8 shows that in fact just the opposite
can be true.
Beyond the algebraic formalism of one-variable substitution in equation (3),
it is worth noting that the term x'(u) represents the "infinitesimal length distortion
factor" involved in the changing from measurement in u to measurement in x.
(See Figure 5.84.) We next attempt to understand how these ideas may be adapted
to the case of multiple integrals.
The Change of Variables Theorem for Double Integrals
Suppose we have a differentiable coordinate transformation from the wu-plane to
the xy-plane. That is,
T: R2 -> R2, T(w, v) = (x(u, v), y(u, v)).
DEFINITION 5.2 The Jacobian of the transformation T, denoted
d(x, y)
d(u, v) '
is the determinant of the derivative matrix DT(u,
v). That is,
dx dx
d(x'y) =detDT(u, u) = det
d(u, v)
du dv
dy dy
du dv
dx dy dx dy
du dv dv du
The notation d(x, y)/d(u, v) for the Jacobian is a historical convenience. The
Jacobian is not a partial derivative, but rather the determinant of the matrix of
partial derivatives. It plays the role of an "infinitesimal area distortion factor"
when changing variables in double integrals, as in the following key result:
5.5 | Change of Variables
y
(f.f)
(8,0)
Figure 5.85 The triangular region
D of Example 9.
THEOREM 5.3 (Change of variables in double integrals) Let D and D*
be elementary regions in (respectively) the xy-plane and the Mu-plane. Suppose
T:R2 — > R2 is a coordinate transformation of class C1 that maps D* onto D
in a one-one fashion. If /: D — > R is any integrable function and we use the
transformation T to make the substitution x = x(u, v), y = y(u, v), then
jj f(x,y)dxdy = j j f(x(u,v),y(u,v))
dp, y)
d(u, v)
du dv.
EXAMPLE 9 We use Theorem 5.3 to calculate the integral
cos(x + 2y) sin(x — y) dx dy
over the triangular region D bounded by the lines y = 0, y = x,andx + 2y = 8 as
shown in Figure 5.85. It is possible to evaluate this integral by using the relatively
straightforward methods of §5.2. However, this would prove to be cumbersome,
so, instead we find a suitable transformation of variables, motivated in this case by
the nature of the integrand. In particular, we let u = x + 2y, v = x — y. Solving
for x and y, we obtain
u + 2v
x = and
y
Therefore,
d(x, y)
d(u, v)
det
Xu Xy
yu yv
det
I 2
3 3
1 _ 1
3 3
Considering the coordinate transformation as a mapping T(w , v) = (x, y) of
the plane, we need to identify a region D* that T maps in a one-one fashion
onto D. To do this, essentially all we need do is to consider the boundaries of D :
y = x
x + 2y = 8
y = 0
x - y = 0
u = 8;
u — v
= 0
v = 0;
v = u.
Hence, one can see that T transforms the region D* shown in Figure 5.86 onto
D. Therefore, applying Theorem 5.3,
Figure 5.86 The effect of the transformation T of Example 9.
Chapter 5 | Multiple Integration
j j cos(x + 2 v) sin(x — y)dx dy = j j cos u sin v
B(x, y)
du dv
d(u, v)
= jj cosw sinv | — 1| dudv
= J J j cos u sin v dv du
Jo Jo
,8
= / | cos u ( — cos v)| JJZq du
Jo
= | / cosw(— cosm + 1) du
Jo
, f*
= 3/ (cos u — cos u) du
Jo
sinwlg — j j(1+cos2m)Jm
sin 8 — [ju + \ sin2M)|pJ
= \ [sin8 - 4
sin
16]
There is another, faster way to calculate the Jacobian, namely, to calculate
d(u, v)/d(x, y) directly from the variable transformation, and then to take reci-
procals. That is, from the equations u = x + 2y, v = x — y, we have
d(u, v)
d(x, y)
det
I'v
Uy
Vy
det
-3.
Consequently, 3(x, y)/d(u, v) = — |, which checks with our previous result.
This method works because if T(h, v) = (x, v), then, under the assumptions of
Theorem 5.3, (u,v) = T_1(jc, y). It follows from the chain rule that
DT-\x,y)= [DT(u,v)]-\
(That is, DT 1 is the inverse matrix of DT. See Exercises 30-38 in § 1 .6 for more
about inverse matrices.) Hence,
^=det[DT-1]=det[(DTr1]
d(u, v)
detOT
EXAMPLE 10 We use Theorem 5.3 to evaluate / fD(x2 - y2) exy dx dy, where
D is the region in the first quadrant bounded by the hyperbolas xy = 1, xy = 4
and the lines y = x, y = x + 2. (See Figure 5.87.)
5.5 | Change of Variables 359
y y=x+2
Figure 5.87 The region D of Example 10.
D*
Figure 5.88 The region D*
corresponding to the region D of
Example 10.
Both the integrand and the region present complications for evaluation. There
would seem to be two natural choices for ways to transform the variables. One
would be
u = x2 — y2 and v = xy,
motivated by the nature of the integrand. However, the region D of integration
will not be easy to describe in terms of this particular choice of wu-coordinates.
Another possible transformation of variables, motivated instead by the shape of
D, is
u = xy and v = y — x .
Now this change of variables would not seem to help much with the integrand,
but, as we shall see, it turns out to be just what we need.
First note that the boundary hyperbolas xy = 1 and xy = 4 correspond re-
spectively, to the lines u = 1 and u = 4; the lines y = x and y = x + 2 correspond
to v = 0 and v = 2. Thus, the region D* in the Mu-plane that corresponds to D
(see Figure 5.88) is
D* = {(u, v) | 1 < u < 4, 0 < v < 2}.
Next, we calculate that the Jacobian of the variable transformation is
9(w, u)
= det
y x
-1 1
= x + y.
d(x, y)
Hence, the Jacobian we require in order to use Theorem 5.3 is
d(x, y) = 1
9(w, v) x + y
Moreover, since we will be working in the first quadrant (where x and y are both
positive), |3(jc, y)/9(w, v)\ = l/(x + y).
Chapter 5 | Multiple Integration
At last we are ready to compute:
jji*2- y2) e*y dx dy = jj (x2 - y2) e*y (v^) du dv
= f2 C (*-?)(* + y)
Jo J i x
+ y
du dv
exy dudv
-u(e4 - el)dv = -y 0 - e )
2
k'
2(e - A
Note that the insertion of the Jacobian in the integrand caused precisely the can-
celation needed to make the evaluation straightforward. We cannot always expect
this to happen, but the lesson here is to be willing to carry through calculations
that may not at first appear to be so easy. ♦
EXAMPLE 11 (Double integrals in polar coordinates) In Example 9, a coor-
dinate transformation was chosen primarily to simplify the integrand of the double
integral. In this example we change variables by using a coordinate system better
suited to the geometry of the region of integration.
For example, suppose that the region D is a disk of radius a :
D={(x,y)\x2 + y2 < a}
= |(x> y) I — V a2 — x2 < y < \l a1 — x2, —a < x < a J .
Then, to integrate any (integrable) function / over D in Cartesian coordinates,
one would write
/ / f(x,y)dxdy= I I f(x,y)dydx.
J Jd J -a J-Ja^l?
Even if it is easy initially to find a partial antiderivative of the integrand, the limits
in the preceding double integral may complicate matters considerably. This is be-
cause the disk is described rather awkwardly by Cartesian coordinates. We know,
however, that it has a much more convenient description in polar coordinates as
{(r, 0) | 0 <r <a,0 <6 < lit).
This suggests that we make the change of variables
(x, y) = T(r, 0) = (r cos#, r sin#),
which is shown in Figure 5.89. (Note that T maps all points of the form (0, 6) to
the origin in the xy-plane and thus, cannot map D* in a one-one fashion onto D.
Nonetheless, the points of D* on which T fails to be one-one fill out a portion
of a line — a one-dimensional locus — and it turns out that it will not affect the
double integral transformation.) The Jacobian for this change of variables is
^ = det
d(r, 6)
cos 9 —rsinO
sin 0 r cos 9
r cos2 9 + r sin2 9 = r.
2k
D*
a
5.5 | Change of Variables
y
361
-~^x2 + y2
A
)
Figure 5.89 T maps the (nonclosed) rectangle D* to
the disk D of radius a.
<
Figure 5.90 The disk of
radius a centered at the
origin.
(0,2)
x2 + y2 = 4
Figure 5.91 The region
D of Example 13.
(Note that r > 0 on D, so \r\ — r.) Thus, using Theorem 5.3, the double integral
can be evaluated by using polar coordinates as follows:
y) dx dy
/a p \J a2—x2
/ f(x,y)dydx
-a J—fa^1
p2n pa
JO JO
f(r cos#, r sin9)r dr d9.
It is evident that the limits of integration of the r0-integrals are substantially
simpler than those in the xy-integral. Of course, the substitution in the integrand
may result in a more complicated expression, but in many situations this will not be
the case. Polar coordinate transformations will prove to be especially convenient
when dealing with regions whose boundaries are parts of circles. ♦
EXAMPLE 12 To see polar coordinates "in action," we calculate the area of a
circle, using double integrals. Once more, let D be the disk of radius a, centered
at the origin as in Figure 5.90. Then we have
Area
r- r- faf V«2— X2 pin pa
ldA= / dydx =
J JD J -a J-Ja2-x2 JO Jo
r dr d9,
following the discussion in Example 1 1 . The last iterated integral is readily eval-
uated as
j jrdrd9 = j (jr2\tyd9 = \a2 dd = \a2(2Tt - 0) = ita2 ,
which indeed agrees with what we already know. If you feel so inclined
compare this calculation with the evaluation of the iterated integral in Cartesian
coordinates. No doubt you'll agree that the use of polar coordinates offers clear
advantages. ♦
EXAMPLE 13 We evaluate the double integral ffD x2 + y2 + ldx dy,
where D is the quarter disk shown in Figure 5.91, using polar coordinates. The
region D of integration is given in Cartesian coordinates by
so that
D = {(x, y) | 0 < y < y^-x2, 0 < x < 2},
f f yjx2 + y2 + \dxdy = f f y 1 x2 + y2 + \dydx
J Jd Jo Jo
Chapter 5 | Multiple Integration
This iterated integral is extremely difficult to evaluate. However, D corresponds
to the polar region
D* = {(r, 9)\0<r <2, 0 < 6> < jt/2}.
Therefore, using Theorem 5.3, we have
J J Vjc2 + y2 + 1 dx dy = ff^>
r2 cos2 9 + r2 sin2 9 + 1 • r dr d6
r2+\rdrd9
= r ' n,
Jo Jo
= / |(r2+l)3/2|_0^
Jo
-L
i(53/2 - \)dd
- !)■
Sketch of a proof of Theorem 5.3 Let (uq, vq) be any point in D* and let
Au = u — uo, Av = v — vo- The coordinate transformation T maps the rectangle
R* inside D* (shown in Figure 5.92) onto the region R inside D in the xy-plane.
(In general, R will not be a rectangle.) Since T is of class C1 , the differentiability
of T (see Definition 3.8 of Chapter 2) implies that the linear approximation
h(u, v) = T(w0, v0) + DT(m0, Wo)
= T(u0, vq) + DT(u0, u0)
U — Mo
v - V0
Au
Av
is a good approximation to T near the point («o , fo)- In particular, h takes the rect-
angle R* onto some parallelogram P that approximates R as shown in Figure 5.93.
We compare the area of R* to that of P.
From Figure 5.93, we see that the rectangle R* is spanned by
0
Am
0
and b
5.5 | Change of Variables
P = h(R*)
DT(u0, v0)b
R = T(R')
T(w0,
h(Mo, v0) ^DT(u0, u0)a
Figure 5.93 The linear approximation h takes the rectangle R* onto a parallelogram P
that approximates R = T(R*).
and the parallelogram P is spanned by the vectors c = DT(mo, t>o)a and d =
DT(wo, wo)b. Hence,
Area of 7?* = ||a x b|| = Am Av,
and thus, by Proposition 5.1,
d(x, y)
Area of P — lie x dll
detDT(w0, Vo)\Au Av
d(u, v)
(u0, v0)
Au Av.
This result gives us some idea how the Jacobian factor arises.
To complete the sketch of the proof, we need a partitioning argument. Partition
D* by subrectangles Rf, . Then we obtain a corresponding partition of D into (not
necessarily rectangular) subregions Rjj = T(7?* ). Let A Ajj denote the area of Rjj .
Let Cij denote the lower left corner of /?*■ and let di; = T(c,7). (See Figure 5.94.)
Then, since / is integrable on D,
/ / f(x,y)dxdy= lim V/(d,7)AAi7.
J JD all Rij-tO *7-f
From the remarks in the preceding paragraph, we know that
d(x,y)
AAjj «a area of parallelogram h(/?* ) =
d(u, v)
(ci7)
Aui Avj.
Figure 5.94 A partition of D* gives rise to a partition of D.
364 Chapter 5 | Multiple Integration
Figure 5.95 The "area element"
d A in rectangular coordinates is
dx dy.
D'
x = r cos 6
y = r sin 0
Figure 5.96 The polar-rectangular transformation takes rectangles in the
r#-plane to wedges of disks in the Jty-plane.
Arclength = rAd
Figure 5.97 An infinitesimal
polar wedge.
Taking limits as all the Rjj tend to zero (i.e., as Am, and Avj approach zero), we
find that
[ [ f(x,y)dxdy = hm Tf (T(c,7))
d(x, y)
d(u, v)
(«y)
All: AVj
If.
f(x(u, v), y(u, u))
d(x, y)
d(u, v)
du dv,
as was to be shown.
Consider again the polar-rectangular coordinate transformation. When we use
Cartesian (rectangular) coordinates to calculate a double integral over a region
D in the plane, then we are subdividing D into "infinitesimal" rectangles having
"area" equal to dx dy. (See Figure 5.95.) On the other hand when we use polar
coordinates to describe this same region, we are subdividing D into infinitesimal
pieces of disks instead. (See Figure 5.96.) These disk wedges arise from trans-
formed rectangles in the r#-plane. One such infinitesimal wedge in the xy-plane
is suggested by Figure 5.97. When AG and Ar are very small, the shape is nearly
rectangular with approximate area (r AO) Ar. Thus, in the limit, we frequently say
d A = dx dy (Cartesian area element)
= r dr d6 (polar area element).
Change of Variables in Triple Integrals
It is not difficult to adapt the previous reasoning to the case of triple integrals. We
omit the details, stating only the main results instead.
DEFINITION 5.4 Let T: R3 -> R3 be a differentiable coordinate transfor-
mation
T(m, v, w) = (x(u, v, w), y(u, v, w), z(u, v, w))
5.5 | Change of Variables
from www-space to xvz-space. The Jacobian of T, denoted
90, y, z)
d(u, v, w)'
is det(DT(w, v, w)). That is,
d(x, y, z)
d(u, v, w)
= det
3x
dx
dx
du
dv
dw
3?
dy
dy_
du
dv
dw
dz
dz
dz
_ du
dv
dw _
In general, given any differentiate coordinate transformation T: R"
the Jacobian is just the determinant of the derivative matrix:
R"
d(xu ...,x„)
d{u\
detDT(M!
3xi
3xi
3xi
du\
du2
du„
dxi
3x2
dx2
det
du\
du2
dun
dxn
dx„
dxn
du\
dU2
dun
THEOREM 5.5 (Change of variables in triple integrals) Let W and W* be
elementary regions in (respectively) xvz-space and wuw-space, and let T: R3 ->
R3 be a coordinate transformation of class C1 that maps W* onto W in a one-one
fashion. If /: W — > R is integrable and we use the transformation T to make the
substitution x = x(u, v, w), y = y(u, v, w), z = z(u, v, w), then
/ / / f{x,y,z)dxdydz
J J Jw
I I Iw*
f(x(u, v, w), y(u, v, w), z(u, v, w))
d(x, y, z)
d(u, v, w)
du dv dw.
(See Figure 5.98.)
In the integral formula of the change of variables theorem (Theorem 5.5), the
Jacobian represents the "volume distortion factor" that occurs when the three-
dimensional region W is subdivided into pieces that are transformed boxes in
Muw-space. (See Figure 5.99.) In other words, the differential volume elements
Chapter 5 | Multiple Integration
Figure 5.98 A three-dimensional transformation T that takes the solid
region W* in uvw-space to the region W in xyz-space.
(i.e., "infinitesimal" pieces of volumes) in xyz- and Muw-coordinates are related
by the formula
dV = dx dy dz =
d(x, y, z)
d(u, v, w)
dudvdw.
Figure 5.99 The volume of the "infinitesimal box" in
uvw-space is du dv dw. The image of this box under T
has volume \d(x, y, z)/d(u, v, w)\ du dv dw.
EXAMPLE 14 (Triple integrals in cylindrical coordinates) When integrat-
ing over solid objects possessing an axis of rotational symmetry, cylindrical
coordinates can be especially helpful. The cylindrical-rectangular coordinate
transformation
' x = r cos 9
y = r sin 6
z = z
has Jacobian
d(x, y, z)
= det
cost'
sinO
0
-r sind
r cos 9
0
= r cos 9 + r sin 9 = r.
d(r, 9, z)
Hence, the formula in Theorem 5.5 becomes
f(r cos6», r sm.9, z)r dr d9 dz.
Ill fix, y, z)dxdydz =
J J Jw J J JV\
5.5 | Change of Variables 367
In particular, we see that the volume element in cylindrical coordinates is
dV = rdrd6dz.
(Recall that the cylindrical coordinate r is usually taken to be nonnegative. Given
this convention, we may omit the absolute value sign in the change of variables
formula.) The geometry behind this volume element is quite plausible: A "dif-
ferential box" in r#z-space is transformed to a portion of a solid cylinder that is
nearly a box itself. (See Figure 5. 100.) ♦
Figure 5.1 00 A "differential box" in rOz-space is mapped to a portion of a
solid cylinder in xyz-space by the cylindrical-rectangular transformation.
EXAMPLE 15 To calculate the volume of a cone of height h and radius a, we
may use Cartesian coordinates, in which case the cone is the solid W bounded by
the surface az = h^/x2 + y2 and the plane z = h, as shown in Figure 5.101. The
volume can be found by calculating the iterated triple integral
/a p ~J a1— x2 p
-a J—Ja^x2 J%
dz dy dx.
We will forgo the details of the evaluation, noting only that trigonometric substi-
tutions are necessary and that they make the resulting computation quite tedious.
az
Shadow in xy-plane
Figure 5.101 The solid cone W of Example 15.
In contrast, since the cone has an axis of rotational symmetry, the use of
cylindrical coordinates should afford us substantially less involved calculations.
Chapter 5 | Multiple Integration
Figure 5.102 The cone of
Example 15 described in
cylindrical coordinates.
Hence, we consider the cone again. (See Figure 5.102.) Note that
h
W =
(r, 9, z)
r < z <h, 0 < r <a, 0 < 9 < lit
Thus, the volume is given by
n n n p 7,71 no nh
dV = / / rdzdrdO.
J J Jw J0 J0 J±r
(Note the order of integration that we chose.) The evaluation of this iterated
integral is exceedingly straightforward; we have
/ / / rdzdrd0= / /
Jo Jo 3\r Jo Jo
-L
-re-
2tt Pa / h
r\h- -r\drd6
h j h
—r 1
2 3a
dO
a xd6
= 2tc I —a'
6
= —a h,
which agrees with what we already know.
EXAMPLE 16 (Triple integrals in spherical coordinates) Ifa solid object has
a center of symmetry, then spherical coordinates can make integration over such
an object more convenient. The spherical-rectangular coordinate transformation
x = p sin <p cos 6
y = p sin cp sin 9
z = p cos cp
has Jacobian
d(x, y, z)
B(p,<p,e)
= det
simp cost
sin<p sin£
coscp
p coscp cos(
p cos cp sin (
— p sin^
-p sin<p sin#
p sin <p cos 0
0
Using cofactor expansion about the last row, this determinant is equal to
cos (p (p2 cos2 9 sin <p cos cp + p2 sin2 9 sin cp cos <p)
+ p sin cp (p cos2 9 sin2 (p + p sin2 1
sin
= p2 cos <p(sin (p cos cp) + p2 sin3 <p
= p2 sin cp (cos2 cp + sin2 (p)
= p2 sincp.
(Under the restriction that 0 < <p < jt, sin^) will always be nonnegative. Hence,
the Jacobian will also be nonnegative.) Therefore, the volume element in spherical
5.5 | Change of Variables
de
dq>
/dp
rdd = psin(p dd
Figure 5.1 03 A differential box in p<p#-space is mapped to a portion of a solid ball
in xyz-space by the spherical-rectangular transformation.
coordinates is
dV = p sin<p dp d<p dO,
and the change of variables formula in Theorem 5.5 becomes
/ / / /(*> y> z)dxdydz
J J Jw
-III
f(x(p, ip, 0), y(p, <p, 0), z(p, (p, 0))p sirup dp dip dO.
The volume element in spherical coordinates makes sense geometrically, because
a differential box in p(p0 -space is transformed to a portion of a solid ball that is
approximated by a box having volume p2 sin q> dp d(p dO. (See Figure 5. 103.) ♦
EXAMPLE 17 The volume of a ball is easy to calculate in spherical coordi-
nates. A solid ball of radius a may be described as
B = {(p, (p,9)\0<p<a,0<(p<7T,0<e< 2jt}.
(See Figure 5 . 1 04.) Hence, we may compute the volume by using the triple integral
fffdV = f!' r r
J J Jb Jo Jo Jo
p sirup dp d(p d9
, . 3
^0 ^0
sin<p dcp dO
- (-coscp\*)d0 = - (-(-l) + l)d0
j Jo j Jo
2a
3 flit
dO
as expected.
EXAMPLE 18 We return to the example of the cone of radius a and height h
and this time, use spherical coordinates to calculate its volume. First, note two
things : (i) that the cone 's lateral surface has the equation (p = tan~ 1 (a / h) in spher-
ical coordinates and (ii) that the planar top having Cartesian equation z = h has
spherical equation p cos cp = h or, equivalently, p = h sec (p. (See Figure 5. 105.)
Chapter 5 | Multiple Integration
For fixed values of the spherical angles tp and 0, the values of p that give
points inside the cone vary from 0 to h sec (p. Any points inside the cone must
have spherical angle <p between 0 and tan~'(a/ h). Finally, by symmetry, 0 can
assume any value between 0 and 2n , Hence, the cone may be described as the set
0 < p < h sec<p, 0 < (p < tan 1 '-, 0 < 0 < 2n \
h
Therefore, we calculate the volume as
"2jt /»tan (a/H) phsecy
pin /'tan \ajn) p
Jo Jo Jo
=/7
^0 ^0
p sin <p dp d<p d0
/h) (h sec<p)3
sin<pd(p dO
^3 r2n ntarTl(a/h)
hi pm pi
= — / / secJ cp siri(p d(p dO
3 Jo Jo
^3 pin ptsarl(alh)
= — / / tmcp sec2 cp dtp dO.
3 Jo Jo
Now, let u = tany> so du = sec2 (p dep. Then the last integral becomes
h3 f2n [a,h h3 f2n 1 /a\2 h3a2 f2lt
— / ududO = — - - ) dO = — - / dO
3 Jo Jo 3 Jo 2 \hj 6h* Jo
3
a2h
-{lit)
Tt
-a h,
as expected.
The use of spherical coordinates in Example 18 is not the most appropri-
ate. We merely include the example so that you can develop some facility with
"thinking spherically." Further practice can be obtained by considering some of
the applications in the next section as well as, of course, some of the exercises.
Summary: Change of Variables Formulas
Change of variables in double integrals:
jj f(x,y)dxdy = j j f(x(u,v),y(u,v
))
d(x, y)
d(u, v)
du dv
Area elements:
dA = dx dy
= rdrd0
djpc, y)
d(u, v)
(Cartesian)
(polar)
dudv (general)
5.5 I Exercises 371
Change of variables in triple integrals:
/ / / /(*> J> z)dxdydz
J J Jw
=fll
f(x(u, v, w), y(u, v, w), z(u, v, w))
d(x, y, z)
d(u, v, w)
du dvdw
Volume elements:
dV = dxdydz (Cartesian)
= r dr d6 dz (cylindrical)
= p2 sin (p dp d<p d9 (spherical)
3(x, y, z)
d(u, v, w)
dudvdw (general)
5.5 Exercises
1. LetT(M, v) = (3m, -v).
(a) Write T(u, v) as A
for a suitable matrix A.
(b) Describe the image D = T(D*), where D* is the
unit square [0, 1] x [0, 1].
2. (a) Let
T(m, v)
u — v u + v
V5 ' V2
How does T transform the unit square D*
[0, 1] x [0, 1]?
(b) Now suppose
T(k, v)
U + V u — V
V2 ' V2
Describe how T transforms D*.
3. If
T(m, v) =
2 3
-1 1
and D* is the parallelogram whose vertices are (0, 0),
(1, 3), (-1, 2), and (0, 5), determine D = T(D*).
4. If D* is the parallelogram whose vertices are (0, 0),
(—1,3), (1, 2), and (0, 5) and D is the parallelogram
whose vertices are (0, 0), (3, 2), (1, —1), and (4, 1),
find a transformation T such that T(Z)*) = D.
5. If T(m, v, w) = (3m — v, u — v + 2w, 5m + 3d — w),
describe how T transforms the unit cube W* =
[0, 1] x [0, 1] x [0, 1].
6. Suppose T(m, d) = (u, uv). Explain (perhaps by us-
ing pictures) how T transforms the unit square D* =
[0, 1] x [0, 1]. Is T one-one on D*l
7. Let T: R3 — >■ R3 be the transformation given by
T(p, <p, 6) = (p sin <p cos 6, p sin<p sin 9, p cos^).
(a) Determine D = T(D*), where D* = [0, 1] x
[O.tt] x [0,2jt].
(b) Determine D = T(D*), where D* = [0, 1] x
[0, 7T/2] x [0, tt/2].
(c) Determine D = T(D*), where D* = [1/2, 1] x
[0, tt/2] x [0, TT/2].
8. This problem concerns the iterated integral
J0 Jy/2
(2x — y)dx dy.
(a) Evaluate this integral and sketch the region D of
integration in the .v v-plane.
(b) Let u = 2x — y and v = y. Find the region D* in
the MD-plane that corresponds to D.
(c) Use the change of variables theorem (Theorem 5.3)
to evaluate the integral by using the substitution
u = 2x — y, v = y.
9. Evaluate the integral
r2 |-(x/2)+l
x/2
by making the substitution u = x, v = 2y — x.
ff
Jo Jxl
x5(2y -x)e(2y~x)2 dydx
Chapter 5 | Multiple Integration
10. Determine the value of
x + y
2y
dA,
where D is the region in R2 enclosed by the lines
y = x/2, y = 0, and x + y = 1 .
1 1 . Evaluate f fD(2x + y)2ex~y dA, where D is the region
enclosed by 2x +y = 1, 2x + y = 4, x — y = — 1,
and x — y = 1 .
12. Evaluate
SL
(2x + y-3)2
dx dy,
D (2y -x + 6)2
where D is the square with vertices (0, 0), (2, 1),
(3, -1), and (1, -2). (Hint: First sketch D and find
the equations of its sides.)
In Exercises 13—1 7, transform the given integral in Cartesian
coordinates to one in polar coordinates and evaluate the polar
integral.
3 dy dx
dy dx
Jo Jo
15. J j (x2 + y2)3/2dA, where D is the disk*2 + v2 <9
/a r^fa1
-a JO
f'f
Jo Jo
dx dy
dy dx
Jx2 + y2
18. Evaluate
11
d ^4"
dA,
where D is the disk of radius 1 with center at (0, 1).
(Be careful when you describe D.)
19. Let D be the region between the square with vertices
(1, 1),(-1, 1),(-1, -1),(1, -1) and the unit disk cen-
tered at the origin. Evaluate
y2 dA.
20. Find the total area enclosed inside the rose r = sin 26 .
(Hint: Sketch the curve and find the area inside a single
leaf.)
21. Let n be a positive integer, and let a be a posi-
tive constant. Calculate the total area inside the rose
r = a cosn8 and show that the value depends only on
a and whether n is even or odd.
22. Find the area of the region inside both of the circles
r = 2a cos 6 and r = 2a sin 9, where a is a positive
constant.
23. Find the area of the region inside the cardioid r =
1 — cos 8 and outside the circle r = 1 .
24. Find the area of the region bounded by the positive
x-axis and the spiral r = 39, 0 < 6 < 2n.
25. Evaluate
cos(x2 + y2)dA,
where D is the shaded region in Figure 5.106.
Arc of a circle
of radius 1
(centered at
origin)
Figure 5.106 The region D of
Exercise 25.
26. Evaluate j j sin (x2 + y2)dA, where D is the region
in the first quadrant bounded by the coordinate axes
and the circles x2 + y2 = 1 and x2 + y2 = 9.
27. Use polar coordinates to evaluate
where D is the unit square [0, 1] x [0, 1].
*3 /.V9-JC2 z-3
dA,
/) fV-r pi
I j
-3 J -^9=3? J \s/x2+y1
ina
/l r-s/\-y2 r4
-1 J-Jl-v2 JO
using cylindrical coordinates.
Jx2 + y
dz dy dx by
29. Evaluate
using cylindrical coordinates.
30. Evaluate
dV
■ dz dx dy by
///.
Ib y/x2 + y2 + z2 + 3 '
where B is the ball of radius 2 centered at the origin.
31. Determine
(x2 + y2+2z2)dV,
where W is the solid cylinder defined by the inequali-
ties x2 + y2 < 4, - 1 < z < 2.
5.6 | Applications of Integration
32. Determine the value of
SSL
dV ', where
>w jx2 + y2
W is the solid region bounded by the plane z = 12 and
the paraboloid z = 2x2 + 2y2 — 6.
33. Find the volume of the region W bounded on top by
z = y/a2 — x2 — y2, on the bottom by the xy-plane,
and on the sides by the cylinder x2 + y2 = b2, where
0 < b < a.
In Exercises 34 and 35, determine the values of the given in-
tegrals, where W is the region bounded by the two spheres
x2 + y2 + z2 = a2 and x2 + y2 + z2 = b2, for 0 < a < b.
«• SSL
dV
35
w jx2 + y2 + z2
■III*
x1 + yL + zz ex
36. Let W denote the solid region in the first octant be-
tween the spheres x2 + y2 + z2 = a2 and x2 + y2 +
z2 = b2, where 0 < a < b. Determine the value of
fffw(x + y + z)dv.
37. Determine the value of fffw z2 dV, where W is the
solid region lying above the cone z = ^3x2 + 3y2 and
inside the sphere x2 + y2 + z2 = 6z.
38. Determine
where W
(l + Jx2 + y2) dV,
(x,y,z)\ y/x2 + y2 <z/2<3\.
39. Find the volume of the region W that represents the
intersection of the solid cylinder x + y2 < 1 and the
solid ellipsoid 2(x2 + y2) + z2 < 10.
40. Find the volume of the solid W that is bounded by
the paraboloid z = 9 — x2 — y2 , the xy-plane, and the
cylinder x2 + y2 = 4.
41. Find
SSL
(2 + x2 + y2)dV,
where W is the region inside the spheres2 + y2 + z2 =
25 and above the plane z = 3.
42. Find the volume of the intersection of the three solid
cylinders
x2 + y2<a2, x2 + z2<a2, and y2+z2<a2.
(Hint: First draw a careful sketch, then note that, by
symmetry, it suffices to calculate the volume of a por-
tion of the intersection. )
Day
°F
Monday
65
Tuesday
63
Wednesday
52
Thursday
51
Friday
45
Saturday
43
Sunday
47
5.6 Applications of Integration
In this section, we explore a variety of settings where double and triple integrals
arise naturally.
Average Value of a Function
Suppose temperatures (shown in the adjacent table) are recorded in Oberlin, Ohio,
during a particular week. From these data, we calculate the average (or mean)
temperature:
65 + 63 + 52 + 51 +45 + 43 + 47
Average temperature
52.3°F.
Of course, this calculation only represents an approximation of the true average
value, since the temperature will vary during each day. To determine the true
average temperature, we need to know the temperature as a function of time for
all instants of time during that one-week period; that is, we consider
Temperature = T(x), x = elapsed time (in days), for 0 < x < 7.
Then a more accurate determination of the average temperature is as an integral:
Average temperature
= 1 fn*
Jo
) dx.
(1)
Since an integral is nothing more than the limit of a sum, it's not hard to see that
the preceding formula is a generalization of the original discrete sum calculation
to the continuous case. (See Figure 5.107.)
374 Chapter 5 | Multiple Integration
80 T
10--
0 1 2 3 4 5 6
Days
Figure 5.1 07 A continuous temperature function T(x)
over the interval [0, 7]. The average temperature for the
week is ^ /Q7 T(x)dx.
Note that
7 = / dx = length of time interval.
Jo
Hence, we may rewrite formula (1) as
Jq T(x)dx
Average temperature =
fo dx
This observation leads us to make the following definitions concerning average
values of functions.
DEFINITION 6.1 (a) Let /: [a, b]^Rbe an integrable function of one
variable. The average (mean) value of / on [a, b] is
i" ft W - faf^dX _ £ /W*
U\«*-u ,1 J Wdx- "length of interval [a, i]-
J a
-f
-a J a
(b) Let /: D c R2 -> R be an integrable function of two variables. The
average value of / on D is
m _IfDfdA_ffDfdA
ffDdA ~ areaof/3'
(c) Let /: W c R3 — > R be an integrable function of three variables. The
average value of / on W is
m IffwfdV ffUfdV
L/Javg
JffwdV volume of W
EXAMPLE 1 Suppose that the "temperature function" for Oberlin during a
week in April is
T(r\ — JIlv7 - Mr6 -I- 1127 r5 _ 2393 4 , 66821 3 _ 45781 2 , 12581 ,
J W — 5040A 180"* 180 A 72 * "1" 720 "* 360 A "1" 210 * "1" UJ'
5.6 | Applications of Integration
(0,1)
(2,0)
Figure 5.108 The triangular
metal plate of Example 2.
where 0 < x < 7. Then the mean temperature for that week would be
ITJavg — 710 / (5040*7
JO
107 6 1 1127 5
180 180
2393 4
72 X
_i_ 66821 3
720 X
1 ( 113 „8
45781 „2 , 12581
360
210
x + 65) c/x
40320
, 66821 4
2880 A
888709
17280
107 7 1 1127 6
1260 1080
2393 5
360
^ + ^ + 65x)\l
1080
51.43°F
EXAMPLE 2 Suppose that the thickness of the triangular metal plate, shown
in Figure 5.108, varies as f(x, y) = xy + 1, where (x, y) are the coordinates of
a point in the plate. The average thickness of the plate is, therefore,
Jo fo~2y(xy + l)dxdy
Average thickness =
fofo 2y dxdy
Note that
1 r2-2y
II
JO JO
dx dy = area of triangular plate = ~(2 • 1) = 1
from elementary geometry. Hence, the average thickness is
fo fo (
xy+l)dxd\ I i\ 7 \|-*=2-2v ,
1 = / (2* y+x)\x=o dy
Jo
= f (\{2-2yfy + (2-2y))dy
Jo
= 2j\y3-2y2+l)dy = 2^- § y3 + y)
EXAMPLE 3 (See also Example 6 of §5.4.) Suppose the temperature inside the
capsule bounded by the paraboloids z = 9 — x2 — y2 and z = 3x2 + 3y2 — 16
varies from point to point as
T(x,y,z) = z(x2 + y2).
We calculate the mean temperature of the capsule.
From Definition 6.1,
[-^lavg —
WwTdV
IffwdV '
The particular iterated integrals we can use for the computation are then
pS/2 ^25/4-x2 ^-x2-y2
/ / / z(x2 + y2)dzdydx
J -5/2 J-^/25/4-A-2 J3x2+3y2-l6
[-^]avg =
/5/2 p y/25/4-x2 p9-x2-y2
I I dzdy dx
-5/2 J-^/25/4-x2 Av2+3j-2-16
Chapter 5 | Multiple Integration
Unfortunately, the calculations involved in evaluating these integrals are rather
tedious.
On the other hand, since the capsule has an axis of rotational symmetry,
cylindrical coordinates can be used to simplify the computations. Note that the
boundary paraboloids have cylindrical equations of z = 9 — r2 and z = 3r2 — 16
and that the shadow of the capsule in the z = 0 plane can be described in polar
coordinates as
{(r,0) | 0 < r < §,0 < 9 < 2n) .
(See Figures 5.109 and 5.1 10.)
{(r,6)\0<r<j,Q<e<2n}
y
z = 3x2 + 3.y2-16
or
z = 3r2- 16
Figure 5.109 The capsule of
Example 3.
Figure 5.110 The shadow of
the capsule in Figure 5.109
in the z = 0 plane.
In addition, the temperature function may be described in cylindrical coordi-
nates as
T(x,y,z) = z(x2 + y2) = zr2.
Hence, we may calculate
[^]avg —
/o* io /2 ii-16 zrl -rdzdr dO
ft ft2 fl/-ibrdzdr de
For the denominator integral,
-lit r5/2 r9-r2
/ / / rdzdr d6 = / r ((9 - r2) - (3r2 - 16)) dr dd
Jo J(i Jlr2-16 Jo Jo
= / / (25r - 4r3) dr de
Jo Jo
-f(i
2n /625 625
r2-r4
5/2
II
-f
Jo
625 625tt
, de = 2jz = .
8 16 7 16 8
5.6 | Applications of Integration
This result agrees with the volume calculation in Example 6 of §5.4, as it should.
For the numerator integral, we compute
p2n p5/2 p9-r2
JO Jo Jir2-U
-2ji r5/2 / .2
zr dzdrd.9
z=9-r2
dr dO
z=3r2-16
Thus,
Jo Jo
= / / ^{(9-r2)2-(3r2-\6f)drde
Jo Jo z
f2n [5/2 r3
= / / — (-8r4 + 78r2 - 175) dr dO
Jo Jo 2
J f2x p5/2
= o/ / (-^7 + 78r5 - I75r3)dr d9
2 Jo Jo
= 2/
_r8 + 13r6 _ 121r4
4
2?r 15625 15625
d6 = TV.
256 256
[^]avg —
■156257T/256 25
625^/8
32
Center of Mass: The Discrete Case
Consider a uniform seesaw with two masses mj and m2 placed on either end. If
we introduce a coordinate system so that the fulcrum of the seesaw is placed at
the origin, then the situation looks something like that shown in Figure 5.111.
Note that x% < 0 < jci. The seesaw balances if
m\X\ + mjX2 = 0.
In this case, the center of mass (or "balance point") of the system is at the origin.
But now suppose m\X\ + m2X2 ^ 0. Then where is the balance point? Let us
denote the coordinate of the balance point by x. Before we find it, we'll introduce
a little terminology. The product m, Xj (in this case, for i = 1, 2) of mass and
position is called the moment of the ith body with respect to the origin of the
coordinate system. The sum m 1X1 + 1112x2 is called the total moment with respect
to the origin. To find the center of mass, we use the following physical principle,
which tells us that a system of several point masses is physically equivalent (in
terms of moments) to a system with a single point mass.
Guiding physical principle. The center of mass is the point such that, if all
the mass of the system were concentrated there, the total moment of the new
system would be the same as that for the original system.
Putting this principle into practice in our situation, we see that total mass M
of our system is mi + m,2. If x is the center of mass, then the guiding principle
tells that
Mx = m\X\ + M2X2-
Chapter 5 | Multiple Integration
m1 m2 m3
X-i
0
Figure 5.112 A system of «
masses distributed on a line.
ml<t
(*i>Vi)
(*3>y3)
Figure 5.113 A system of «
masses in R2.
That is, the total moment of the new (concentrated) system is the same as the total
moment of the original system. Hence,
m\X\ + 1H2X2
X =
If we have a system of n masses distributed along a (coordinatized) line, then
the same reasoning may be applied. (See Figure 5.1 12.) We have
total moment ni\X\ + 1712X2 + • • • + mnxn Yl"=i mixi
total mass
mi + m2 H h m„
(2)
Now we move to two and three dimensions. Suppose, first, that we have n
particles (or bodies) arranged in the plane as in Figure 5.1 13. Then there are two
moments to consider:
n
Total moment with respect to the y-axis = m,*, ,
1=1
and
Total moment with respect to the x-axis = m, -y,- .
1=1
(Admittedly, this terminology may seem confusing at first. The idea is that the
moment measures how the system balances with respect to the coordinate axes.
It is the x-coordinate — not the y-coordinate — that measures position relative to
the y-axis. Similarly, the y-coordinate measures position relative to the x-axis.)
The guiding principle tells us that the center of mass is the point (x , y) such that,
if all the mass of the system were concentrated there, then the new system would
have the same total moments as the original system. That is, if M = ^ m;, then
Mx = rriiXi
(i.e., the moment with respect to the y-axis of the new system equals the moment
with respect to the y-axis of the original system) and
My = y^m,y,-.
Thus, we have shown the following:
Discrete center of mass in R2. Given a system of n point masses m\,
m2, . . . , m„ at positions
(xi,yi), (x2, y2) (x„,yn) inR2,
the coordinates (x , y) of the center of mass are
5.6 | Applications of Integration 379
For particles arranged in three dimensions, little more is needed than adding
an additional coordinate. (See Figure 5.1 14.)
Discrete center of mass in R . Given a system of n point masses m\,
ni2, ■ . ■ , mn at positions
(xuyi,zi), (x2, y2, Zi), ■ ■ - , (x„,y„,z„) inR3,
the coordinates (x, y, z) of the center of mass are given by
*= , y=^' — , and z=^' '■ (4)
The numerators of the fractions in (4) are the moments with respect to the
coordinate planes. Thus, for example, the sum Yl"=i mix> is me total moment
with respect to the yz-plane.
By definition, moments of physical systems are additive. That is, the total
moment of a system is the sum of the moments of its constituent pieces. However,
it is by no means the case that a coordinate of the center of mass of a system is
the sum of the coordinates of the centers of mass of its pieces. This additivity
property makes the study of moments important in its own right.
Center of Mass: The Continuous Case
Now, we turn our attention to physical systems where mass is distributed in
a continuous fashion throughout the system rather than at only finitely many
isolated points.
To begin with the one-dimensional case, suppose we have a straight wire
placed on a coordinate axis between points x = a and x = b as shown in
Figure 5.115. Moreover, suppose that the mass of this wire is distributed according
to some continuous density function <5 (x ) . We seek the coordinate x that represents
the center of mass, or "balance point," of the wire.
xn=b
Figure 5.1 1 5 A "coordinatized" wire.
The mass of the segment between jc;_i and
Xi is approximately 8(x*)Axj.
Imagine breaking the wire into n small pieces. Since the density is continuous,
it will be nearly constant on each small piece. Thus, for i = 1 , . . . , n, the mass m,
of each piece is approximately 8(x*)Ax, ■ , where A the length of
each segment of wire, and x* is any number in the subinterval [jc,_i , #,•]. Hence,
the total mass is
n n
M = J2'ni ~ ^2s(x*)Axi,
i=l i=l
and the total moment with respect to the origin is approximately
xf 8(x*)Ax,.
approx. approx.
position mass
mi
- y
X
Figure 5.1 1 4 A discrete system
of masses in R3.
J- — I 1 1 — H-
d — Xq X\ X-j ■ ■ ■ X; _ i X;
380 Chapter 5 | Multiple Integration
Of course, these results can be used to provide an approximation of the coordi-
nate x of the center of mass. For an exact result, however, we let all the pieces
of wire become "infmitesimally small"; that is, we take limits of the foregoing
approximating sums as all the Ax, 's tend to zero. Such limits give us integrals,
and we may reasonably define our terms as follows:
Continuous center of mass in R. For a wire located along the x-axis
between x = a and x = b with continuous density per unit length <5(x):
f
Total mass = / 8(x)dx
Total moment = / xS(x)dx. (5)
J a
total moment f xS(x)dx
Center of mass x =
total mass f* 8(x)dx
Compare the formulas in (3) with those in (5). Instead of a sum of masses
and a sum of products of mass and position, we have an integral of "infinitesimal
mass" (the <5(x) dx term) and an integral of infinitesimal mass times position.
EXAMPLE 4 Suppose that a wire is located between x = — 1 and x = 1 along
a coordinate line and has density S(x) = x2 + 1. Using the formulas in (5), we
compute that the center of mass has coordinate
Figure 5.116 A lamina
depicted as a region D in
the xy-plane with density
function S.
f\x{x2+\)dx _ {jx^+jx^ti
f\(x2+l)dx (ix3+x)|lj
4 = 0.
3
This makes sense, since this wire has a symmetric density pattern with respect to
the origin (i.e., <5(x) = S(—x)). ♦
The analogous situation in two dimensions is that of a lamina or flat plate of
finite extent and continuously varying density S(x, y). (See Figure 5.116.) Using
reasoning similar to that used to obtain the formulas in (5), we make the following
definition for the coordinates (x, y) of the center of mass of the lamina:
Continuous center of mass in R2. For a lamina represented by the region
D in the xy-plane with continuous density per unit area 8(x, y):
x =
total moment with respect to y-axis ffDx S(x, y)dA
' JfDS(x,y)dA ;
ffDyS(x,y)dA
total mass
total moment with respect to x-axis
total mass
(6)
ffDS(x,y)dA ■
5.6 | Applications of Integration
Roughly, the term S(x, y)dA represents the mass of an "infinitesimal two-dimen-
sional" piece of the lamina and the various double integrals the limiting sums of
such masses or their corresponding moments.
EXAMPLE 5 We wish to find the center of mass of a lamina represented by the
region D in R2 whose boundary consists of portions of the parabola y = x2 and
the line y = 9 and whose density varies as S(x, y) = x2 + y. (See Figure 5.117.)
First, note that this lamina is symmetric with respect to the y-axis and that, in
addition, the density function has a similar symmetry because 8 (x, y) = S(—x, y).
We may conclude from these two observations that the center of mass must occur
along the y-axis (i.e., that x. = 0). Using the formulas in (6) and noting that the
lamina is represented by an elementary region of type 1 ,
y =
JfDyS(x,y)dA _ j% f^yjx2 + y)dy dx
ffDS(x,y)dA : : /\ • y)dydx
For the denominator integral, we compute
x + y) dy dx =
dx
For the numerator,
9
y(x2 + y)dy dx =
dx
x6 x6\l
1 dx
2 3 )\
3
11664
7
Hence,
11664/7 _ 45
1296/5 ~ T
6.43.
This answer is quite plausible, since the density of the lamina increases with y,
and so we should expect the center of mass to be closer to y = 9 than to y = 0.
♦
382 Chapter 5 | Multiple Integration
We may modify the two-dimensional formulas to produce three-dimensional
ones.
Continuous center of mass in R3. Given a solid W whose density per
unit volume varies continuously as 8(x, y, z), we compute the coordinates
(x, y, z) of the center of mass of W using the following quotients of triple
integrals:
total moment with respect to yz-plane fffwx8(x,y,z)dV
total mass
fffwS(x,y,z)dV '
y
total moment with respect to xz-plane fffwy Kxi y, z)dV
total mass fffw 8(x, y, z)dV
total moment with respect to x y-plane fffwz S(x, y,z)dV
total mass fffwS(x,y,z)dV
-; (7)
(0,0,3)
Plane
x + y + z =3
(3,0, 0)
Figure 5.118
Example 6.
The tetrahedron of
In (7) we may think of the term S(x, y, z)dV as representing the mass of an
"infinitesimal three-dimensional" piece of W. Then the triple integrals are the
limiting sums of masses or moments of such pieces.
EXAMPLE 6 Consider the solid tetrahedron W with vertices at (0, 0, 0),
(3, 0, 0), (0, 3, 0), and (0, 0, 3). Suppose the mass density at the point (x, y, z)
inside the tetrahedron is 8(x, y,z) = x + y + z+ 1. We calculate the resulting
center of mass. (See Figure 5.118.)
First, note that the position of the tetrahedron in space and the density function
are both such that the roles of x, y, and z may be interchanged freely. Hence, the
coordinates (x, y, z) of the center of mass must satisfy x — y — z. Therefore, we
may reduce the number of calculations required.
The tetrahedron is a type 4 elementary region in space. Thus, we may calculate
the total mass M of W, using the following iterated integral:
l-x r3-x-y
M
if I
Jo Jo Jo
fo fo \
Jo Jo
= f m-
Jo
(x + y + z + l)dzdy dx
Z=3—x—y
(x + y + \)z +
1T2
2X
y -xy
dy dx
2 y2) dy dx
x2)y-\{\+x)y2-\y%
1-2
2
=3-.i
dx
The total moment with respect to the xy-plane is given by
O r3-x i-3-x-y
ITI
Jo Jo Jo
z(x + y + z + 1) dz dy dx
5.6 | Applications of Integration
•3 ,3-, / 2 z3
(x + y+l)- + j
If
Jo Jo
/T"<
Jo Jo
=3-x-y
dy dx
2 2X + +
+ xy+ \x2y + \y2 + \xy2 + iy3) dy dx
Jo
117 27,
15 „2 _ 1 v3
4 A 6A
lx4)dx=459
Hence,
x = y = z =
459
40
117
51
65
24
0.7846.
40 ■
If an object is uniform, in the sense that it has constant density, then one uses
the term centroid to refer to the center of mass of that object. Suppose the object
is a solid region W in R3. Then, if the density <5 is a constant k, the equations for
the coordinates (x,y,z) may be deduced from those in (7). For the x-coordinate,
we have
x =
fffwxS(x,y,z)dV = fffwkxdV
fffw8(x,y,z)dV fffwkdV
fffwxdV _ 1
fffwdV volume of W
1 1 Iw
xdV.
Similarly,
y =
1
volume of W
f f fy
ydV and z=
volume of W
ffl
zdV.
In particular, the constant density S plays no role in the calculation of the
centroid, only the geometry of W. (Note: Completely analogous statements can
be made in the case of centroids of laminas in R2.)
Figure 5.1 19 The cone of
Example 7.
EXAMPLE 7 We compute the centroid of a cone of radius a and height h . (See
Figure 5.119.)
By symmetry, x = y = 0. Moreover, we know that the volume of the cone is
(ir/3)a2h. Thus, the z-coordinate of the centroid is
= —111
na2h J J Jw
zdV.
This triple integral is most readily evaluated by using cylindrical coordinates.
(See Example 15 of §5.5.) The lateral surface of the cone is given by z = \r, so
we calculate
^ /,2jr pa
= — 7T- I I I zrdzdrd0 =
nazh J0 Jo J'ir
Tca2h
jta2h2\ 3
I = -h
4/4
after a straightforward evaluation. Hence, the centroid of the cone is located at
(0,0, \h). ♦
384 Chapter 5 | Multiple Integration
Moments of Inertia
Let W be a rigid solid body in space. As we have seen, the moment integral
with respect to the xy-plane is Mxy = fffwz 8(x, y, z)dV — that is, the inte-
gral of the product of the position relative to a reference plane (in this case the
xy-plane) and the density of the solid. This integral can be considered to mea-
sure the ease with which W can be displaced perpendicularly from the reference
plane.
Now, consider spinning W about a fixed axis (which may or may not pass
through W). The moment of inertia / (or second moment — the moment integral
mentioned in the preceding paragraph is sometimes called the first moment) is
a measure of the ease with which W can be made to spin about the given axis.
Specifically, / is the integral of the product of the density at a point in W and the
square of the distance from that point to a fixed axis; that is,
1 1 Iw
dl8{x,y,z)dV, (8)
w
where d is the distance from (x, y, z) 6 W to the specified axis.
When the axes of rotation are the coordinate axes in R3, we have
/v = moment of inertia about the x-axis = / / / (y + z )8(x, y, z) dV;
'w
ffl
Iy = moment of inertia about the y-axis = / / / (x2 + z2) S(x, y, z) d V;
J J Jw
I- = moment of inertia about the z-axis = / / / (x2 + y2)S(x, y, z)dV.
J J Jw
Figure 5.120 The box of
Example 8.
EXAMPLE 8 Let W be a solid box of uniform density S and dimensions a, b,
and c. If W is situated symmetrically with respect to the coordinate axes as shown
in Figure 5.120, we compute the moments of inertia with respect to these axes.
Note, first, that W may be described as
W = \(x,y,z)
a a b be c
- < x < -, — < y < — < z < -
2 ~ ~ 2 2~ y ~ 2 2 ~ "2
Hence, the moment of inertia about the x-axis is
/c/2 rb/2 pa /2 pc/2 pb/2
/ / (v2 + z2)Sdxdydz= / / (y2 + z2)Sadydz
-c/2 J-b/2 J -a/2 J -c/2 J-b/2
fc/2 / 3 \ y=bl2 fc /2 /b3 \
=SaLS^+zy),---Jz=SaLM+bz)
+ bzz dz
,'b^c bc^
Sal 1
12 12
Sabc
12
'-(b2 + c2).
5.6 | Applications of Integration
By permuting the roles of x, y, and z (and the corresponding constants a, b, and
c), we see that
Sabc 0 i Sabc , ,
ly = —{a2 + c2) and L = —{a2 + b2).
Therefore, if a > b > c (as in Figure 5.120), it follows that Ix < Iy < /,. This
result may be confirmed by the observation that rotations about the axis parallel
to the longest side of the box are easiest to effect in that the same torque applied
about each axis will cause the most rapid rotation to occur about the axis through
the longest dimension. A related fact is regularly exploited by figure skaters who
pull their arms in close to their bodies, thereby reducing their moments of inertia
and speeding up their spins. ♦
EXAMPLE 9 Let W be the solid bounded by the cone z = 2^/x2 + y2 and the
plane z = 4 shown in Figure 5.121. Assume that the density of material inside W
varies as 8(x, y, z) = 5 — z. Let us calculate the moment of inertia Iz about the
z-axis.
Given the geometry of the situation, it is easiest to work in cylindrical coor-
dinates, in which case the cone is given by the cylindrical equation z = 2r. Thus,
we have
L = f [ [ (x2 + y2)S(x,y,z)dV = f f f r2(5 - z)rdzdrd6
J J Jw Jo Jo Jlr
= / r3 (5z ~ \z2)\Z~2r drM = / / (12r3 " 10r4 + 2r5) drde
Jo Jo Jo Jo
f2n , i s 1 /r\ i r=2 f2lT 16 32;r
Jo Jo
Recall that the center of mass of a solid object of total mass M is the point
such that if all the mass M were concentrated there, the (first) moment would
remain the same. An analogous idea may be defined in the context of moments of
inertia. The radius of gyration of a solid with respect to an axis is the distance
r from that axis that we should locate a point of mass M so that it has the same
moment of inertia / (with respect to the axis) as the original solid does. More
concisely, the radius of gyration r is defined by the equation
EXAMPLE 10 We determine the radius of gyration with respect to the z-axis
of the cone described in Example 9. Hence, we compute
386 Chapter 5 | Multiple Integration
From Example 9, L = 32n/3. We determine the total mass M of the cone as
follows:
M= / / (5 - z)rdzdrd6 = / (12 - lOr + 2r2)r dr d9
J0 J0 Jlr Jo Jo
=r((*i-^+Mi^*=r^=f-
Thus,
' Y 32^/3
5.6 Exercises
1 . The local grocery store receives a shipment of 75 cases
of cat food every month. The inventory of cat food (i.e.,
the number of cases of cat food on hand as a function
of days) is given by I(x) = 75 cos(jrx/15) + 80.
(a) What is the average daily inventory over a month?
(b) If the cost of storing a case is 2 cents per day,
determine the average daily holding cost over the
month.
2. Find the average value of f(x, y) = sin2 x cos2 y over
R = [0,2jt] x [0,4tt].
3. Find the average value of f(x, y) = e2x+y over the tri-
angular region whose vertices are (0, 0), (1, 0), and
(0, 1).
4. Find the average value of g(x, y, z) = ez over the unit
ball given by
B = {(x,y,z) | *2 + y2 + z2 < 1}.
5. Suppose that the temperature at a point in the cube
W = [-1, 1] x [-1, 1] x [-1, 1]
varies in proportion to the square of the point's distance
from the origin.
(a) What is the average temperature of the cube?
(b) Describe the set of points in the cube where the
temperature is equal to the average temperature.
6. Let D be the region between the square with ver-
tices (1, 1), (-1, 1), (-1, -1), (1, -1) and the unit
disk centered at the origin. Find the average value of
f(x, y) = x2 + y2 on D.
7. Let W be the region in R3 between the cube with
vertices (1, 1, 1), (-1, 1, 1), (-1, -1, 1), (1, -1, 1),
(1,1,-1), (-1,1,-1), (-1,-1,-1), (1,-1,-1)
and the unit ball centered at the origin. Find the av-
erage value of f(x, y, z) = x2 + y2 + z2 on W.
8. Suppose that you commute every day to work by sub-
way. You walk to the same subway station, which is
served by two subway lines, both stopping near where
you work. During rush hour, each subway line sends
trains to arrive at the stop every 6 minutes, but the dis-
patchers begin the schedules at random times. What is
the average time you expect to wait for a subway train?
(Hint: Model the waiting time for the two subway lines
by using a point (x, y) in the square [0, 6] x [0, 6].)
9. Repeat Exercise 8 in the case that the subway stop is
serviced by three subway lines (each with trains arriv-
ing every 6 minutes), rather than two.
10. Find the center of mass of the region bounded by the
parabola y = 8 — 2x2 and the *-axis
(a) if the density S is constant;
(b) if the density S = 3y.
11. Find the centroid of a semicircular plate. (Hint: Judi-
cious use of a suitable coordinate system might help.)
1 2. Find the center of mass of a plate that is shaped like the
region between y = x2 and y = 2x, where the density
varies as 1 + x + y.
13. Find the center of mass of a lamina shaped like the
region
{(*, y) | 0 < y < *Jx, 0 < x < 9},
where the density varies as xy.
1 4. Find the centroid of the region bounded by the cardioid
given in polar coordinates by the equation r =
1 — sin 9. (Hint: Think carefully.)
15. Find the centroid of the lamina described in polar
coordinates as
{(r,9) | 0 < r < 4cos0, 0 < 6 < tt/3}.
5.6 I Exercises 387
1 6. Find the center of mass of the lamina described in polar
coordinates as
l(r,6) | 0 < r < 3, 0 <6 < tt/4},
where the density of the lamina varies as S(r, 0) =
4-r.
17. Find the center of mass of the region inside the car-
dioid given in polar coordinates as r = 1 + cos 8, and
whose density varies as S(r, 6) = r.
1 8. Find the centroid of the tetrahedron whose vertices are
at (0, 0, 0), (1, 0, 0), (0, 2, 0), and (0, 0, 3).
19. A solid is bounded below by z = 3y2, above by the
plane 2 = 3, and on the ends by the planes x = — 1
and x = 2.
(a) Find the centroid of this solid.
(b) Now assume that the density of the solid is given
by S = z + x2. Find the center of mass of the
solid.
20. Determine the centroid of the region bounded above
by the sphere x2 + y2 + z2 = 1 8 and below by the
paraboloid 3z = x2 + y2 .
21 . Find the centroid of the solid, capsule-shaped region
bounded by the paraboloids z = 3x2 + 3y2 — 16 and
z = 9-x2-y2.
22. Find the centroid of the "ice cream cone" shown in
Figure 5.122.
z
Sphere: x2 + y2 + z2 = 25
x
Figure 5.122 The ice cream cone solid
of Exercise 22.
23. Find the centroid of the solid shaped as one-eighth of
a solid ball of radius a. (Hint: Model the solid as the
first octant portion of a ball of radius a with center at
the origin. )
24. Find the center of mass of a solid cylindrical peg of
radius a and height h whose mass density at a point
in the peg varies as the square of the distance of that
point from the top of the cylinder.
25. (a) Find the moment of inertia about the coordinate
axes of a solid, homogeneous tetrahedron whose
vertices are located at (0,0,0), (1, 0, 0), (0, 1, 0),
and(0, 0, 1).
(b) What are the radii of gyration about the coordinate
axes?
26. Consider the solid cube W = [0, 2] x [0, 2] x [0, 2].
Find the moments of inertia and the radii of gyration
about the coordinate axes if the density of the cube is
S(x, y, z) = x + y + z + 1.
27. A solid is bounded by the paraboloid z = x2 + y2 and
the plane z = 9. Find the moment of inertia and radius
of gyration about the z-axis if
(a) the density is S(x, y, z) = 2z;
(b) the density is S(x, y, z) = *Jx2 + y2.
28. Find the moment of inertia and radius of gyration about
the z-axis of a solid ball of radius a, centered at the ori-
gin, if
(a) the density S is constant;
(b) S(x,y,z) = x2 + y2 + z2;
(c) S(x,y,z) = x2 + y2.
We can find the moment of inertia of a lamina in the plane
with density S(x, y)by considering the lamina to be a flat plate
sitting in the xy -plane in R3. Then, for example, the distance
of a point (x, y) in the lamina to the x-axis is given by \y\, the
distance to the y-axis is given by \x\, and the distance to the
z-axis (or the origin) is given by *J x2 + y2. (See Figure 5.123.)
Using these ideas, find the specified moments of inertia of the
laminas given in Exercises 29—31.
z
29. The moment of inertia Ix about the x-axis of the
lamina that has the shape bounded by the graph of
y = x2 +2 and the line y = 3, and whose density
varies as S(x, y) = x2 + 1.
30. The moment of inertia /, about the z-axis of the lam-
ina shaped as the rectangle [0, 2] x [0, 1], and whose
density varies as S(x, y) = 1 + y.
388 Chapter 5 | Multiple Integration
31 . The moment of inertia about the line y = 3 of the lam-
ina shaped as the disk
{(x,y)\x2 + y2 <4},
and whose density varies as S(x, y) = x2.
The gravitational field between a mass M concentrated at
the point (x, y, z) and a mass m concentrated at the point
(xo, Jo, Zo) is
p_ GMm[(x - x0)i + (y - ypjj + (z - z0)k]
[(x - x0)2 + (y - y0)2 + (z - za)2fl2
The gravitational potential V ofF is
_ GMm
y/(x -x0)2 + (y- y0)2 + (z - Z0)2 '
(We have seen in §3.3 that F = -VV.J Now suppose that, in-
stead of a point mass M, we have a solid region W of density
S(x, y, z) and total mass M. Then the gravitational potential of
W acting on the point mass m may be found by looking at "in-
finitesimal "point masses dm = S(x, y,z)dV and adding (via
integration) their individual potentials. That is, the potential
ofW is
V(x0, y0, z0)
err GmS(x,y,z)dv
J J Jw J{x - x0)2 + (y - y0)2 + (z - z0)2 '
In Exercises 32-34, Let W be the region between two con-
centric spheres of radii a < b, centered at the origin. (See
Figure 5.124.) Assume that W has total mass M and constant
density 8. The object of the following exercises is to compute
the gravitational potential V(xo, yo, Zo) of W on a mass m
concentrated at (xo, yo, Zo)- Note that, by the spherical sym-
metry, there is no loss of generality in taking (xq, yo, Zo) equal
to (0, 0, r). So, in particular, r is the distance from the point
mass m to the center ofW.
Figure 5.124 The spherical
shell of Exercises 32-34.
32. Show that if r > b, then V(0, 0, r) = -GMm/r. This
is exactly the same gravitational potential as if all the
mass M of W were concentrated at the origin. This is a
key result of Newtonian mechanics. (Hint: Use spher-
ical coordinates and integrate with respect to <p before
integrating with respect to p.)
33. Show that if r < a, then there is no gravitational force.
(Hint: Show that V(0, 0, r) is actually independent of r.
Then relate the gravitational potential to gravitational
force. As in Exercise 32, use spherical coordinates and
integrate with respect to (p before integrating with re-
spect to p.)
34. (a) Find V(0, 0, r) if a < r < b.
(b) Relate your answer in part (a) to the results of Ex-
ercises 32 and 33.
5.7 Numerical Approximations of Multiple
Integrals (optional)
Suppose that / is a continuous function defined on some bounded region D in the
plane. If D is a rectangle or an elementary region, then Theorem 2. 1 0 enables us to
calculate ffDf dA as an iterated integral. However, the "partial antiderivatives"
in the iterated integral may turn out to be impossible to determine in practice.
Nonetheless, an approximate value of ffD f dA may be entirely acceptable for
some purposes. In this section, we discuss how we can adapt numerical meth-
ods for approximating definite integrals of functions of a single variable to give
techniques for approximating double integrals.
Numerical Methods for Definite Integrals
Let us review two familiar techniques for approximating the value of fa f(x) dx.
The first of these is the trapezoidal rule. We begin by partitioning the interval
[a, b] into n equal subintervals. Thus, we set
b — a
Ax = , a = xo < x\ < . . . < xn = b, where Xj = a + i Ax.
n
5.7 | Numerical Approximations of Multiple Integrals (optional)
The trapezoidal rule approximation T„ is
Tn = ^ U(a) + 2 fix,) +■■■+ 2/(x»_i) + f(b)]
(1)
It is obtained by approximating the function / with n linear functions that pass
through the pairs of points (x,_i , /(x,_i)) and (x,- , /(*,•)) for r = 1, . . . , n. Then
the (net) area under the curve y = f(x) between x = a andx = b is approximated
by the sum of the (net) areas under the graphs of the linear functions. When /
is nonnegative, the approximating areas are those of trapezoids — thus, the name
for the method. (See Figure 5.125.)
The key theoretical result concerning the trapezoidal rule is given by the
following.
Ax
~2~
n-l
/(a) + 2 £/(*,) + /(&)
THEOREM 7.1 Given a function / that is integrable on [a, b], we have
/'
J a
f{x)dx = Tn + E„,
where
Ax
~1~
n-l
f(a) + 2j2f(xi) + f(t>)
and E„ denotes the error involved in using Tn to approximate the value of the
definite integral. In addition, if / is of class C2 on [a, b], then there is some
number f in (a , b) such that
F —
(Ax)2/"(0 =
12
12n2
In particular, Theorem 7.1 shows that if /" is bounded that is, if \f"(x)\ < M
for all x in [a, b], then \En\ < (b — a)3 M/(l2n2). This inequality is very useful
in estimating the accuracy of the approximation.
EXAMPLE 1 We approximate f0 sin(x2) dx using the trapezoidal rule with
n = 4. Thus, we have
1 - 0
Ax = = 0.25,
4
so that, using (1), we have
I
1 0 25
sin(x2)Jx R« T4 = — [sin0 + 2sin(0.252) + 2sin(0.52) + 2sin(0.752)
+ sinl]
0.25
2
0.25
[0 + 0.12492 + 0.49481 + 1.06661 + 0.84147]
[2.52780] = 0.31598.
Chapter 5 | Multiple Integration
Note that the second derivative of sin(x2) is 2 cos (x2) — Ax2 sin(x2) so that, for
0 < x < 1,
|2cos(x2)-4x2sin(x2)| < 2| cos(x2)| + 4x2| sin(x2)| < 2 + 4 = 6.
Hence, using Theorem 7.1,
(1 - 0)36 1
|£4| < - V- = — = 0.03125.
1 1 " 12 -42 32
Thus, the true value ofthe integral must be between 0.3 1598 - 0.03125 = 0.2847
(rounding to four decimal places) and 0.31598 + 0.03125 = 0.3473. Of course,
a more accurate approximation may be obtained by taking a finer partition (i.e.,
a larger value for n). ♦
Another numerical technique for approximating the value of f(x)dx with
which you may be familiar is Simpson's rule. As with the trapezoidal rule, we
partition the interval [a, b] into equal subintervals, only now we require that the
number of subintervals be even, which we will write as 2n. Thus, we take
b — a
Ax = , a = xo < X\ < . . . < X2n = b, where x,- = a + / Ax.
2n
Then the Simpson's rule approximation Sin is
Ax
Sm = — If (a) + 4/(xi) + 2/(x2) + 4/(x3) + • ■ ■ + 2/(x2„_2)
+ 4/(x2n_0 + /(&)]
n — 1 n
f(a) + 2 £ /(x2() + 4 £ /(x2l_.) + f(b)
Ax
(2)
EXAMPLE 2 We approximate the value of f0 sin(x2) dx of Example 1 using
Simpson's rule with n = 2 (i.e., four subintervals). As before, we have
1 - 0
Ax = = 0.25,
4
so (2) gives
■! o 25
sin (x2) dx R» 54 = — [sin 0 + 4 sin (0.252) + 2 sin (0.52) + 4 sin (0.752)
/
Jo
+ sin 1]
0.25
3
0.25
[0 + 0.24984 + 0.49481 + 2.13321 + 0.84147]
[3.71933] = 0.30994.
3
Note that this value is in the range predicted by the trapezoidal rule. In fact, it is
a more accurate approximation to the value of the definite integral. ♦
Simpson's rule is obtained by approximating the function / with n quadratic
functions that pass through triples of points (x2,_2, /(x2(_2)), (x2;_i, /fe-i)),
(x2i , f(xii)) for i = 1 , . . . , n. As with the trapezoidal rule, we have the following
summary result.
5.7 | Numerical Approximations of Multiple Integrals (optional)
THEOREM 7.2 Given a function / that is integrable on [a, b], we have
b
f(x)dx = S2n + E2n,
where
n— 1 n
f(a) + 2 £ /(xa) + 4 J] /(x2i_0 + /(*) ,
i=i i=i
and Ein denotes the error involved in using S^i to approximate the value of the
definite integral. In addition, if / is of class C4 on [a, b], then there is some
number £ in (a , b) such that
E, - -b~a(Ax)4 f^>(f) - - (fc - a) f(4)m
2" " 180 { ' 1 (;) ~ 2880/24 1 {;)-
L
Sin =
Ax
EXAMPLE 3 Consider (x3 + 3x2) dx. We compare the trapezoidal rule
and Simpson's rule approximations with 4 subintervals. We thus have Ax =
(3 - l)/4 = 0.5 and
0.5
T4 = — [4 + 2(10.125 + 20 + 34.375) + 54] = 46.75;
0.5
S4 = — [4 + 4(10.125) + 2(20) + 4(34.375) + 54] = 46.
Note that
(x3 + 3x2) dx = (I*4 + x3) \\ = (f + 27) - (i + 1) = 46,
so Simpson's rule agrees with the exact answer. This should not be a surprise,
since for f(x) = x3 + 3x2, we have that f{4\x) is identically zero, which means
that the error term £4 for Simpson's rule must be zero. ♦
Approximating Double Integrals over Rectangles
Now let / be a function of two variables that is continuous on the rectangle
R = [a, b] x [c, d] in R2. We adapt the previous ideas to provide methods for
approximating the value of the double integral JjRfdA.
Because we assume that / is continuous on R, Fubini's theorem applies to
give
/ / f(x,y)dA= f f f(x,y)dydx. (3)
J J R J a Jc
Next we partition the x-interval [a, b] into m equal subintervals. Thus,
b — a
Ax = , a = xo < x\ < . . . < xm = b, where x, ■ = a + / Ax.
m
Similarly, we partition the y-interval [c, d] into n equal subintervals; hence,
d — c
Ay = , c = yo < v\ < . . . < y„ = d, where y;- = c + j Ay.
n
In the inner integral fd f(x, y) dy of the iterated integral in (3), the variable x is
held constant. Therefore, we may approximate this integral using the trapezoidal
Chapter 5 i Multiple Integration
rule of Theorem 7.1. We obtain
f{x,y)dy
Ay
n-l
/(ac,c) + 2^)/(x,y;) + /(x,d)
7 = 1
Next, integrate each function of x appearing on the right side, so that
-b pel
J a J c
Ay
2
y)dydx &
r-b n — 1 pb r-b
J f{x, c)dx + 2 I fix^yj)dx+ I fix,d)dx
J a ,-_ i J a J a
(4)
Now use the trapezoidal rule again on each integral appearing on the right. This
means that, for j = 0, . . . , n,
f
J (I
fix, yj)dx
Ax
f(atyj) + 2j2f(Xi,yj) + m yj)
(5)
Putting (4) and (5) together, we obtain
p b pel
/ / /(*,
J a J c
y) dy dx
Ay Ax
m— 1
n-l
Ax
fia,c) + 2 YJfi.xi,c) + fib, c)
m— 1
f(a.y,) • 2j./|v,.v/) • /(/>. V;)
+
7=1
Ax
i=i
fia,d) + 2YJf(xi,d) + fib,d)
Therefore, the trapezoidal rule approximation T„hn to ffRf dA is
T
1 m.n
Ax Ay
-l
fia,c) + 2j2f(xi,c) + fQ>,c)
( = 1
n— 1 m — 1
n— 1 n— 1 m— 1 ft— 1
+ 2 J] /(a, y,) + 4££ /(x, , yj) + 2 J] fib, yj) (6)
7 = 1 7 = 1 «=1 7 = 1
1-1
+ fia,d) + 2YJf{xi,d) + fib,d)
i=i
The expression appearing in (6) is not too memorable as it stands. However,
we may interpret it as follows:
Ax Ay -r-^ ^-^
Tm,„ = —— 2^ yj)>
j=0 1=0
5.7 | Numerical Approximations of Multiple Integrals (optional)
where
1 if (xi , yj) is one of the four vertices of R ;
2 if (xi , yj) is a point on an edge of R, but not a vertex;
4 if (x, , yj) is a point in the interior of R.
EXAMPLE 4 We approximate the value of L jx {xy + 3x)dydx with 7^2.
Thus, the ^-interval [3, 6] is partitioned using Ajc = (6 — 3)/3 = 1 and the y-
interval [1, 2] is partitioned using Ay = (2 — l)/2 = 0.5. See Figure 5.126 for
the rectangle [3,6] x [1,2] with partition points marked.
Hence, we have
(xy + 3x) dy dx «
1(0.5)
[12 + 24 + 30+ 15
1
+ 2(16 + 20 + 27 + 20 + 25 + 13.5) + 4(18 + 22.5)]
-(486) = 60.75.
y
2 --
1.5
1 +
15
13.5,
12
20
25
30
18
22.5
27
•
•
16
20
24
H 1 1 h
H h
Figure 5.126 The rectangle
[3, 6] x [1, 2] of Example 4 with
partition points marked with the
values of f(x, y) = xy + 3x.
For comparison, we calculate the iterated integral directly:
2
dx
y=l
f6 9 9 2
= / -x dx = -x
J3 2 4
Note that T3 2 gives the exact answer in this case. ♦
In the derivation above of the trapezoidal rule we assumed that the func-
tion / was continuous. Nonetheless, we may use formula (6) to approximate
ffRf dA whenever / is integrable on R. However, in order to make estimates
of the accuracy of trapezoidal approximations, we must assume more about /, as
the following result (whose proof involves use of the intermediate value theorem
and the mean value theorem for integrals) indicates.
if;
(xy + 3x)dy dx
= f(
xy-
2
3xy
243
= 60.75.
394 Chapter 5 | Multiple Integration
THEOREM 7.3 (TRAPEZOIDAL RULE FOR DOUBLE INTEGRALS OVER RECTANGLES)
Given a function / that is integrable on the rectangle R = [a , b] x [c, d], we have
//.
fix, y)dA = Tm%n + Em>n,
R
where T„u„ is given by (6) and Emj, denotes the error involved in using T„un to
approximate the value of the double integral. Moreover, if / is of class C2 on R,
then there exist points (fi , rj\) and (£2, Vi) in R such that
Em,n = _(fc-°X^-c) [(A*)2/^, ,x) + (Av)2/yyfe, ift)] •
EXAMPLE 5 For /(*, y) = xy + 3x, we have that d2f/dx2 and d2f/dy2 are
both identically zero. Hence, Theorem 7.3 implies that the trapezoidal rule ap-
proximation is exact, as we already observed in Example 4. ♦
In a manner entirely analogous to our derivation of the trapezoidal rule, we
may also produce a Simpson's rule approximation to the value of the double
integral ffR f dA. As in the case of Simpson's rule for approximating single-
variable definite integrals, we partition both the x- and y-intervals into even
numbers of equal subintervals. Thus, we take
b — a
Ax = , a = xq < x\ < . . . < X2m = b, where x,- = a + / Ax,
2m
and
d — c
Ay = , c = yo < y\ < . . . < ym = d, where y7- = c + j Ay.
2n
The resulting Simpson's rule approximation S2m,2n is given by the expression
Ax Ay
m— 1 m
fia, c) + 2j2 f(*2i ,c) + 4J2 f(.X2i-u c) + fib, c)
(=1 (=1
— 1 n— 1 m— 1 n— 1 m
+ 2 f(a, yij) f(*»> *y) + 8 E E
;=1 7=1 i=l ;=1 t=l
n-1
+ 2J2fib,y2j) (7)
7 = 1
+ fia, yij-x) + /(**. yy-i)
7=1 ;=1 i=l
n m n
+ 16 E E /c^-i. ^-i) + 4 E ^-0
7=1 i=l 7=1
m — 1 m
+ /(fl, J) + 2 J] /(x2„ rf) + 4£ f(x2i-i,d) + fib, d)
/=! i=i
5.7 | Numerical Approximations of Multiple Integrals (optional)
Just as in the case of the trapezoidal rule for double integrals over rectangles, the
intermediate value theorem and the mean value theorem for integrals provide the
following result.
THEOREM 7.4 (Simpson's rule for double integrals over rectangles)
Let / be a function that is integrable on the rectangle R = [a, b] x [c, d]. Then
f(x, y)dA = S2,„,2n + ^2m,2n>
where S2m,2n is given by (7) and E2m,2„ denotes the error involved in using S%m^n
to approximate the value of the double integral. Moreover, if / is of class C4 on
R, then there exist points (£i , r]\) and (£2, m) in R such that
E2m,2n = ~(b~a^ KAX)4 + (Ay)Vvvvv(£2, m)] ■
EXAMPLE 6 We compare approximations to the value of /0°'5 /„' ex+y dydx
using both the trapezoidal rule and Simpson's rule with four subintervals in
each of the x- and y-intervals. Thus, the x-interval [0, 0.5] is partitioned using
Ax = (0.5 — 0)/4 = 0.125 and the y-interval [0, 1] is partitioned using Ay =
(1 -0)/4 = 0.25.
The trapezoidal rule gives
T4A = (Q-125X°-25> yO+0 + e0+0.5 + el+0 + e0.5+l + 2 ^0+0.25 + e0+0.5
_|_ ^0+0.75 _j_ e0.125+0 _j_ ^0.25+0 _|_ ^0.375+0 _|_ ^0.125+1 _|_ ^0.25+1
_|_ ^0.375+1 _|_ ^0.5+0.25 _j_ e0.5+0.5 _j_ ^0.5+0.75^
0.125+0.25 | „0.125+0. 5 _|_ g0. 125+0.75 _|_ g0.25+0.25
_|_ g0.375+0.5 + ^0.375+0.75^ j
+ 4(e"
, ^0. 25+0. 75 , g0.37
1
= (143.608854) = 1.121944.
128
Note that d2/dx2 (ex+y) = d2/dy2 (ex+y) = ex+y . Theorem 7.3 says that there
exist points rji) and (£2, V2) in the rectangle [0, 0.5] x [0, 1] such that
Eaa = _(0-5-0)(l -0) ^(0_125)2efl+1J1 + (0.25)V2+"2] .
On the rectangle [0, 0.5] x [0, 1], the smallest possible value of ex+y is e0+0 = 1
and the largest possible value is e0 5+l = eLS. Hence, we must have
_(Ogl) j-(0 125)2ei.5 + (0.25)2^-5] < e4A < [(0.125)2 + (0.25)2]
or
-0.0145888 < £44 < -0.003255.
Chapter 5 | Multiple Integration
|e0+0.25 + ^0+0.75 + g0.125+0 + ^0.125+1 + ^0.25+0.5
g0.375+0 . g0.375+l . g0.5+0.25 , ^0.5+0.75-,
Hence, the true value of the double integral lies between 1.121944 — 0.0145888 =
1.10735 and 1.121944 - 0.003255 = 1.11869.
On the other hand, Simpson's rule gives
54,4 = (0-125^(0-25) + + eH0 + ^0.5+1
+ 2 (e°+0-5 + e°-25+0 + e°-25+1 + e°-5+0-5)
+ 4(e
" 5)
_|_ g ^0.125+0.5 _|_ ^0.25+0.25 _|_ g0.25+0.75 _|_ g0.375+0.5^
_|_ ! g ^0.125+0.25 _|_ ^0.125+0.75 _|_ ^0.375+0.25 _|_ ^0.375+0.75^
= —(321.036910) = 1.114711.
288
In this case, we note that d4/dx4 (ex+y) = d4/dy4 (ex+y) = ex+y, so, as before,
the minimum and maximum values of these partial derivatives on [0, 0.5] x [0, 1]
are, respectively, 1 and e . Therefore, Theorem 7.4 implies that
[(°-125)V-5 + (0-25)V-5] < £4.4 < [(0.125)4 + (0.25)4]
or
-0.0000516688 < £4,4 < -0.0000115289.
Hence, the true value of the double integral lies between 1.114711 —
0.0000516688 = 1.11466 and 1.114711 - 0.0000115289 = 1.1147.
For comparison, we may calculate the iterated integral exactly:
/ / ex+ydydx=\ (ex+y)f dx = / (ex+l - ex) dx
Jo Jo Jo ' Jo
= (ex+l - ex)\°Q5 = ehS - e0-5 - e + \ ^ 1.114686.
Thus, we see that Simpson's rule gives a highly accurate approximation with a
very coarse partition. ♦
Approximating Double Integrals over Elementary Regions
We can modify the methods for approximating double integrals over rectangles
to approximate double integrals over more general regions. Suppose first that D
is an elementary region of type 1 ; that is,
D = {{x, y) 6 R2 I y(x) < y < S(x), a <x <b}.
5.7 | Numerical Approximations of Multiple Integrals (optional)
Then, if / is continuous on D, Theorem 2.10 tells us that
Iy(x)
fdA= / f(x,y)dydx.
J JD J a Jy(x)
To approximate this iterated integral with a version of, say, the trapezoidal rule,
we need to partition the region D in a reasonable way. We do so as follows. First,
we partition the x -interval [a, b] in the usual way:
b — a
Ax = , a = xo < x\ < . . . < xm = b, where xl ■ = a + iAx.
m
Now, for a fixed x in [a, b], we partition the corresponding y-interval [y(x), S(x)]
into n equal subintervals:
. . , <5(x) - y(x)
Ay(x) = ,
Y(x) = yo < y\ < ■ ■ ■ < yn = §(x), where yj(x) = y(x) + JAy(x).
Note that now Ay and the partition numbers yo , . . . , y„ are all functions of x . (See
Figure 5.127.) Then, by applying the trapezoidal rule first to the inner integral
and then the outer one, we obtain
pb pS(x)
J a J y(x )
fix, y)dy dx
L
» Ay(x)
n-l
f(x, yix)) + 2^/(x, yjix)) + fix, Six))
dx
Ax
~1~
Ay id) ^— \ Ay(x, )
-——fia, y(a)) + 2 ^ — —fixi, y(x,))
Ay(fe)
n-l
——fia, yjia)) + 2 ^ — - — y7- (*,■))
+ -±2f(b,yjib))
+
+
2
Ay (a)
2
Ay(&)
m— 1
/(a, 5(a)) + 2 J] /(*,-, «(*,))
398 Chapter 5 | Multiple Integration
The trapezoidal rule approximation is, thus,
T =
Ax Ay (a)
4
-i-i
f(a, y(a)) + 2 J] f(a, yj(a)) + f(a, 8(a))
AxAy(xi)
n-l
2f(Xi , y(xt)) + f(*i , (8)
7 = 1
+ 2/(xi,<5(xi))
+
AxAy(Z?)
n-l
y(6)) + 2 J] /(ft, yj(b)) + /(&, 5(6))
EXAMPLE
4
3.5
3
2.5
2
2 2.1 2.2
Figure 5.1 28 The partitioned
region D of Example 7.
7 We approximate /22'2 J"2* (x3 + y2) 6?y dx by T2i4. We have
2.2-2
Ax = = 0.1, so xo = 2, xi = 2.1, X2 = 2.2
and, therefore, that
4-2
Ay(2) = = 0.5,
so yo(x0) = 2, y1(x0) = 2.5, yz(xo) = 3, y3(xo) = 3.5, y4(x0) = 4,
4.2-2.1
Ay(2.1) = = 0.525,
so y0(xi) = 2.1, yi(xi) = 2.625, y2(xi) = 3.15, y3(xj) = 3.675,
y4Oi) = 4.2,
4.4-2.2
Ay(2.2) = = 0.55,
so yo(x2) = 2.2, yi(x2) = 2.75, y2(x2) = 3.3, yi(x2) = 3.85,
y4(x2) = 4.4.
(See Figure 5.128.) Thus,
T2A = (°"1)4(Q,5) [(23 + 22) + 2 ((23 + 2.52) + (23 + 32) + (23 + 3.52))
+ (23 + 42)]
+ (0,1)^'525) [2(2. 13 + 2.12) + 4(2. 13 + 2.6252) + 4(2. 13 + 3.152)
+ 4(2. 13 + 3.6752) + 2(2. 13 + 4.22)]
+ (0-1^0-55^ [(2.23 + 2.22) + 2(2.23 + 2.752) + 2(2.23 + 3.32)
+ 2(2.23 + 3.852) + 2(2.23 + 4.42)]
= 0.0125(139) + 0.02625(156.7755) + 0.01375(175.934) = 8.271949375.
5.7 | Numerical Approximations of Multiple Integrals (optional)
In this case, the exact answer is
.2.2 plx p2.2
p2.2 p2x n2.2 n2.2
J J (x3 + .v2) dy dx = J (x3y + ±y3) |^ dx = J (x4 + |x3) dx
= 8.23886.
I2-2 _ 1/T)5
12 5
(2.25 - 25) + i (2.24 - 24)
If Z) is an elementary region of type 2, that is,
D = {(x, y) e R2 | a(y) < x < fi(y), c < y < d},
then we may similarly approximate ffDf dA by first partitioning the y-interval
[c, d] using
d — c
Ay = , c = y0 < yi < . . . < y„ = d, where y ; = c + jAy,
n
and then, for a fixed y in [c, d], by partitioning the corresponding x-interval
[a(y), /5(y)] into m equal subintervals:
£00 -
Ajc(y) =
77?
a(y) = x0 < Xi < . . . < xm = /3(y), where x,(y) = a(y) + i Ax(y).
In doing so, we obtain a counterpart formula to that of (8), namely,
T
± in
Ax(c)Ay
4
(-1
| g Ax(yy-)Ay
/(a(c), c) + 2 /(*/(c), c) + /(/3(c), c)
m-l
2/(«(y,).yD • 4j]/(.v,(y, ). y;))
7 = 1
1=1
+ 2f(P(yj),yj)
(9)
+
Ax (J) Ay
m-l
/(a(d), d) + 2j2 f(xi(d), d) + f (J3(d), d)
i=i
Of course, we may also adapt Simpson's rule for use in the case of double
integrals over elementary regions. Moreover, we may derive similar methods for
approximating triple integrals as well. In practice, however, other methods are
often used that lend themselves to computer implementation.
One such alternative technique is known as the Monte Carlo method. It is
based on a result called the mean value theorem for double integrals: If / is
a continuous function of two variables and D is a bounded and connected (i.e.,
one-piece) region in R2, then there is a point P e D such that
f(x, y)dA = f(P) ■ area of D.
400 Chapter 5 | Multiple Integration
5.7 Exercises
Note that it follows that
ffnfdA
area ot D
so that, by Definition 6.1, we have that f(P) = [f]w%, the average (mean) value
of / on D. The Monte Carlo method for approximating ffDf dA is to select n
points P\ , . . . , Pn in D at random and compute the average value / of / on just
these points:
Then / will approximate [/]avg and, hence,
- (area of D)
fdA~ (area of D)f = K- T f(P,).
n ' 4
i=\
If the area of D is in turn difficult to determine exactly, it, too, may be estimated
by situating D inside a rectangle R and selecting m points at random in R. If r
denotes the fraction of these points lying in D, then area of D « r ■ (area of R).
In an analogous manner, we may give Monte Carlo approximations for triple
integrals of functions of three variables defined over solid regions in space.
In Exercises 1—6, (a) use the trapezoidal rule approximation
72,3 to estimate the values of the given integrals, and (b) com-
pare your results with the exact answers.
*3.1 /-2.1
l.i r0.6
pi. I nl.\
1. J J (x2-6y2) dydx
/■3.3 z-3.3
2. / / xy2 dy dx
13 J3
-2.2 fl.6
dy dx
n
J [~Jx + <fy) dy dx
/l.l /.0.6
J ex+2ydydx
6. / / x cos y dy dx
J0 Jjt/6
In Exercises 7—12, (a) use the approximation S2jfrom Simp-
son s rule to estimate the values of the given integrals, and (b)
compare your results with the exact answers.
-OA /.0.3
/U.l />U.J
/ {y4 - xy2) dy dx
-o.i Jo
pOA p2
JO Jl
1 +X'
dy dx
Ii Jo
Jo Jn
?x+2},dvdx
-7i/4 rn/2
10. / / sin2x cos3y dy dx
rr/4
px/4 rn/4
11.1 / sin (x + y) dy dx
Jo Jo
1.1 rx/4
Ii Jo
ex cos v dy dx
13. In Chapter 7 we will see that the area of the portion of
the graph of f(x, y) for (x, y) in a region D in R2 is
given by the double integral / fD J f 2 + f2 + 1 dA.
(a) Set up an appropriate iterated integral to compute
the surface area of the portion of the paraboloid
z = 4 - x2 - 3y2 where (x, y) e [0, 1] x [0, 1].
(b) Use the trapezoidal rule approximation 74 4 to es-
timate the surface area.
14. Concerning the iterated integral
f\'5 f\.4ln(2x + y) dydx:
(a) Calculate the trapezoidal rale approximation T^.
(b) Use Theorem 7.3 to estimate the error in your ap-
proximation in part (a).
(c) Calculate the Simpson's rule approximation 52,4.
True/False Exercises for Chapter 5 401
(d) Use Theorem 7.4 to estimate the error in your ap-
proximation in part (a).
15. Without either evaluating or estimating the integral
fi 4 Jo s ln(xy)dy dx, which approximation is more
accurate: 74 4 or 52,2? Explain your answer.
1 6. Suppose that the trapezoidal rule is used to estimate the
value of /0°'2 /^qj exl+2y dydx. Determine the small-
est value of n so that the resulting approximation T„t„
is accurate to within 10~4 of the actual value of the
integral.
17. Consider J0°'3 J0°'4 ex~y dy dx.
(a) If the trapezoidal rule approximation Tn „ is used
to estimate the value of this integral, what is
the smallest value of n so that the resulting
approximation is accurate to within 10~5 of the
actual value?
(b) If the Simpson's rule approximation S2,,, i„ is used
to estimate the value of this integral, what is the
smallest value of n so that the resulting approx-
imation is accurate to within 10~5 of the actual
value?
18. Concerning the iterated integral JqJq(3x +
5y)dy dx:
(a) Calculate the trapezoidal rule approximation 7/2,2.
(b) Compare your result in part (a) with the exact
answer.
(c) Use Theorem 7.3 to explain your results in parts
(a) and (b).
19. Consider the iterated integral j\ f^2 x3y3 dy dx:
(a) Calculate the Simpson's rule approximation £2,2-
(b) Compare your result in part (a) with the exact an-
swer.
(c) Use Theorem 7.4 to explain your results in parts
(a) and (b).
In Exercises 20-25, (a) use the approximation T3 3 from the
trapezoidal rule to estimate the values of the given integrals.
20. J J (x 3 + 2y2) dy dx
/*jr/4 pcosx
21 . / / (2x cos y + sin2 x) dy dx
Jo J sinx
,.0.3 fix
22. J J (xy - x2) dy dx
23. / / dydx
Jo Jo v 1 — y2
24. J j sin x dx dy (Note the order of integration.)
/1.6 />2y
J In (xy)dxdy
26. In this problem, you will develop another way to think
about the trapezoidal rule approximation given in equa-
tion (6).
(a) Let L denote a general linear function of two vari-
ables, that is, L(x, y) = Ax + By + C, where A,
B, and C are constants. Set R = [a, b] x [c, d].
Show that
/ / LdA = (area of W)(average of the values of
•> JR L taken at the four vertices of/?).
(Note that this gives an exact expression for the
double integral.)
(b) Suppose that / is any function of two variables that
is integrable on R. Show that the trapezoidal rule
approximation Ti \ to ffRf dA is
7^1 = (area of /?)(average of the values off
taken at the four vertices of R).
(c) Now let Ax = (b — a)/m, Ay = (d — c)/n,
and, for i = 1 , . . . , m, j = 1, . . . , n, let /?,•_,• =
[Xi-i, x{\ x [y,_i, y,], where xt = a + iAx and
y/ = c + j Ay. Then we have
///<" = ££ // fdA-
Use 7iti to approximate each integral ffR fdA
and sum the results to obtain the formula for T„un
given by equation (6).
True/False Exercises for Chapter 5
1. Every rectangle in R2 may be denoted [a, b] x [c, d].
f1/2 f2 3 2
. I y sin {jix ) dy dx
2. If / is a continuous function and f(x, y) > 0 on a
region D in R2, then the volume of the solid in R3 r2 Z"1/2
under the graph of the surface z = f(x, y) and above = J i Jo ^ s^n ^nx ^ dx
the region D in the xy-plane is / / f(x,y)dA.
402 Chapter 5 | Multiple Integration
/ / 3 dy dx = J J 3 dx dy
Jo Jo Jo Jo
10 Jo
-2 ry
5- / / f(x,y)dxdy= I I f(x, y)dy dx for
Jo Jo Jo Jo
all continuous functions /.
6. 0 < j J sin(x4 + y4)dx dy < jt, where D denotes
the unit disk {(x, y) \ x1 + y2 < 1}.
7. f [ sfx3 + ydydx = [ [ ^x3 + y dx dy.
Jo Jo Jo Jo
8. J x2ex+y dydx=(^J x2ex dx\(j^ ey dy^j.
9. The region in R2 bounded by the graphs of y = x3
and y = ~Jx is a type 3 elementary region in the
plane.
10. The region in R2 bounded by the graphs of y = sinx,
y = cosx,x = u j A, andx = 5 jr/4 is a type 2 elemen-
tary region in the plane.
1 1 . The region in R3 bounded by the graphs of y2 — x2 —
z2 = 1 and 9x2 + 4z2 = 36 is a type 3 elementary
region in space.
12. j j (y3 + l)dxdy gives the area of the region D,
where D = {(x, y) | (x - 2)2 + 3y2 < 5}.
' // <
angle with vertices (-1,0), (1, 0), (0, 1).
13. / / (y2 sin (x3) + 3) dx dy = 3 /2, where D is the tri-
(x + y3+z5)dV = 0.
[-2,2]x[-l,l]x[-3,3]
. [ [ I (x + z)dV = 0.
/[-2,2]x[0,l]x[-l,l]
SSL
■SSL
l l l[-2,2]
Jo Jv/4 J-
x[0,l]x[-l,l]
4 n^y/2 f2
(y + z)dV = 0.
sin yz dz dx dy = 0.
18. / / / (y -x2)dzdydx = 0.
J-l J-JT^x2 J-^/l-x2-/
19. If T(n, v) = (2m — v, u + 3d), then the area of the im-
age/) = T(D*)ofthe unit square/)* = [0, 1] x [0, 1]
is 7 square units.
20. If T(m, v) = (v — u,3u + 2v), then the area of the im-
age D = T(D*) of the rectangle D* = [0, 3] x [0, 2]
is 5 square units.
21
77
JO Jy
2 r5-2y
12
e2x y cos(x — 3y) dx dy
10 nu/2
I f
JO Ju
-5e" cos vdvdu.
22. If D is the disk {(x, y) | x2 + y2 < 9}, then
jj v/9-x2-y2dA = 36jr.
/>2jt f>2 p4
23. The iterated integral / / / dzdrdO represents
Jo Jo J2r
the volume enclosed by the cone of height 4 and
radius 2.
24. The iterated integral
ffZS
J-2 J-V^x2 JO
{2 + Jx2 + y2)dzdy dx
is given by an equivalent integral in cylindrical coor-
dinates as
2m p2 /.79-r2
pill pi r>
Jo Jo Jo
(r2 + 2r)dz dr d6.
10 Jo Jo
25. The iterated integral
fV2=^ n^4-x2-y2
/ / / 7x2 + y2 +z2 +5dzdydx
J-s/2.Jo J^/x2+y2
is given by an equivalent integral in spherical coordi-
nates as
/ / / Pv P2 + 5 sin<pdpd<pd8.
Jo Jo Jo
26. The average value of the function /(x,y) = x2y
over the semicircular region D = {(x, y) | 0 < y <
V4 — x2} is given by
1 rn/2 r2
or by
1 f' f
- / / r4sin6»cos26Uri/6»
x Jo Jo
— f [ r4 sin9 cos2 9 drd6.
X Jn/2 Jo
27. The center of mass of a lamina represented by the tri-
angle with vertices (2, 0), (0, 1), (0, —1), and whose
density varies as S(x, y) = (x2 + l)cosy, has coordi-
nates given by
fo fu-2)/2 (*3 + *) cos y dy dx
r2 rv-*)/ifY2
y = 0.
Jo !(x-2)/2^x2 + U cos y dy dx
28. The centroid of a cone of radius a, height h, with axis
the z-axis and vertex at (0, 0, h) is (0, 0,
29. The center of mass of the solid cylinder of radius a,
height h , with axis the z-axis whose density at any point
Miscellaneous Exercises for Chapter 5 403
30. The integral / / / p5 sin2</5 siri(pdp dtp d8 repre-
J J Jw
varies as ed , where d is the square of the distance from
the point to the z-axis is (0, 0, z), where
f" f'' zrerl dzdrdO
— \h. W with density z, expressed in spherical coordinates.
r2n ra rh .2 , , ,„ sents the moment of inertia about the z-axis of a solid
Jo Jo Jo zre dzdrdd
fo2" So So re* dzdr de
Miscellaneous Exercises for Chapter 5
1. Let B be the ball of radius 3; that is,
B = {(x,y,z)\x2 + y2+z2 < 9}.
Without resorting to any explicit calculation of an iter-
ated integral, determine the value of the triple integral
fffB(z3 +2)dV by using geometry and symmetry
considerations.
2. Let W denote half of the solid ball of radius 2; that is,
W = {(x, y, z) | x1 + y2 + z2 < 4, z > 0}.
Without resorting to explicit calculation of an iterated
integral, determine the value of the triple integral
SSL
(x3+y-3)dV.
(Hint: Use geometry and symmetry.)
3. Let W be the solid region in R3 with x > 0 that is
bounded by the three surfaces z = 9 — x2, z = 2x2 +
y2, and x = 0.
(a) Set up, but do not evaluate, two different (but
equivalent) iterated integrals that both give the
vzlueof fffw 3 dV.
(b) Use a computer algebra system to find the value
of Jffw3dV and to check for consistency in your
answers in part (a).
4. Suppose that / is continuous on the rectangle R =
[a, b] x [c, d]. For (x, y) G (a, b) x (c, d), we define
F(x,y)-
ff
J a J c
f(x', y')dy'dx'
Use Fubini's theorem to show that d2F/dxdy =
d2F/dydx. This provides an alternative proof of the
equality of mixed partials. (Hint: Write
where
F(x, y)
*(*', y)
f g(x',
J a
y)dx'
j" f(x',y')dy'.
Then dF/dx and d2F/dydx may be calculated using
the fundamental theorem of calculus. Then use Fubini's
theorem to find dF/dy and d2 F/dxdy.)
5. Convert the following cylindrical integral to equivalent
iterated integrals in (a) Cartesian coordinates and (b)
spherical coordinates:
p2n p\ p -\
Jo Jo Jo
r dz dr d6.
lo Jo Jo
Evaluate the easiest of the three iterated integrals.
6. The volume of a solid is given by the iterated integral
Jo Jo J-^/i
dz dx dy.
(a) Sketch the solid and also describe it by giving equa-
tions for the surfaces that form its boundary.
(b) Express the volume as an iterated integral in cylin-
drical coordinates. Determine the volume.
7. Calculate the volume of a cube having edge length a
by integrating in cylindrical coordinates. (Hint: Put the
center of the cube at the origin.)
8. Calculate the volume of a cube having edge length a
by integrating in spherical coordinates.
9. Determine
cos
2y
X + V
dA,
where D is the triangular region bounded by the coor-
dinate axes and the line x + y = 1 .
10. Evaluate
/>6 />1— 2y
Jo J-2v
y\x + 2y)2e(x+2y? dxdy
by making a suitable change of variables.
1 1 . Find the area enclosed by the ellipse E given by the
equation
~ + ~b~2
1
a
in the following way:
(a) First, write the area as the value of an appropriate
iterated integral in Cartesian coordinates. Do not
evaluate this integral.
(b) Next, scale the variables by letting x = ax,
y = by. To what region E* in the iy-plane does
404 Chapter 5 , Multiple Integration
the ellipse E correspond? Rewrite the xy-integral
in part (a) as an xy-integral.
(c) Finally, use polar coordinates to transform the xy-
integral and thereby show that the area inside the
original ellipse is nab.
1 2. This problem concerns the rotated ellipse E with equa-
tion ttx1 + I4xy + 10y2 = 9.
(a) Let u = 2x — y, v = x + y and rewrite the equa-
tion for E in the form
+
b-
l,
where a and b are positive constants to be
determined.
(b) Use an appropriate change of variables and the re-
sult of part (c) of Exercise 1 1 to find the area en-
closed by E .
13. Consider the ellipse E with equation
5x2 + 6xy + 5y2 = 4. Let u = x — y, v = x + y and
follow the steps of Exercise 12.
1 4. Imitate the techniques of Exercise 1 1 to find the volume
enclosed by the ellipsoid E given by the equation
2 2 2
x y Z
h 1
a2 b2 c2
1.
15. Evaluate
xy
d y — x
dA,
where D is the region in the first quadrant bounded
by the hyperbolas x
y
1, x — y = 4 and
the ellipses x2/4 + y2 = 1, x2/\6 + y2/4 = 1. (Hint:
Sketch the region D, and use it to make an appropriate
change of variables.)
16. Evaluate
(x2 + y V
dA,
where D is the region in the first quadrant bounded
by the hyperbolas x2 — y2 = 1, x2 — y2 = 9, xy = 1,
xy = 4.
17. Evaluate
dA,
where D is the region bounded by xy = 1, xy = 4,
y = l, y = 2.
1 8. (a) Generalizing the notions of double and triple inte-
grals, develop a definition of the "quadruple inte-
gral" // ffw fdVofa function f(x, y, z, w) over
a four-dimensional region W in R4.
mi
(b) Use your definition in part (a) to calculate
(x + 2y + 3z-4w)dV,
r
where W is the four-dimensional box
W = {(x, y, z, w) | 0 < x < 2, -1 < y < 3,
0 < z < 4, -2 < w < 2}.
19. (a) Set up, but do not evaluate, a quadruple iterated in-
tegral that computes the four-dimensional volume
of the four-dimensional ball of radius a :
B = {(x, y, z, w) | x2 + y2 + z2 + w2 < a2}.
(b) Use a computer algebra system to give a formula
for the volume.
(c) Use a computer algebra system to give for-
mulas for the n-dimensional volume of the
M-dimensional ball
B = {{xx,x2, . . . , Xn) I x\ + x\ H Yx2n< a2}
in the cases where n = 5, 6. Is there any pattern to
your answers?
In Exercises 20—23, you will give a general expression for the
n-dimensional volume of the n-dimensional ball
B = {(xx,x2,...,xn) | x2+x2 + ---x2 < a2}.
Let V„(a) denote this n-dimensional volume. Let C„ denote the
n-dimensional volume of the unit ball U; that is, C„ = V„(l).
20. By scaling the variables x\,x2, ■ ■ ■ ,xn and using a
change of variables formula analogous to those in The-
orems 5.3 and 5.5, show that V„(a) = Cnan.
21. (a) Consider points in B of the form (jci, x2, 0, 0).
Show that the coordinates x\, x2 describe points
(in R2) lying in the disk of radius a.
(b) In R", let the polar coordinates r and 9 replace
the Cartesian coordinates X\ and x2. Argue that
the points in B lying over a particular point (r, 6)
in the disk described in part (a) must fill out an
(n — 2)-dimensional ball of radius V 'a2 — r2.
(c) Use part (b) to show that the w-dimensional volume
of B is given by
f2jT
■de.
f I Vn.2(Va2-r2)rdr,
Jo Jo
22. Use the previous two exercises to establish the recur-
sive formula
(2jta2\
Vn(a) = Vn-2{a).
23. (a) Show that V\{a) = 2a and V2(a) = it a2. (These
are familiar facts.)
Miscellaneous Exercises for Chapter 5 405
(b) Use part (a) and the previous problem to show that
Vn(a) :
7T"'2
(n/2)!
2(n+l)/2jr(«-l)/2
if n is even
a" if n is odd
where the double factorial n\\ = n(n — 2)(« —
4) • • • 3 • 1 (i.e., the product of all odd integers
from 1 to n).
24. A spherical shell with inner radius 3 cm and outer
radius 4 cm has a mass density that varies as
0.12c?2 g/cm3, where d denotes the distance (in cen-
timeters) from a point in the shell to the center of the
shell.
(a) Determine the total mass of the shell.
(b) Will the shell float in water? (Note: The density
of water is 1 g/cm3. To answer this question, you
need to determine the average density of the shell.)
(c) Suppose that the shell has a small hole so that the
core of the shell fills with water. Now will it float?
25. A dome is shaped as a hemisphere. If a pole whose
length is the average height of the dome is to be in-
stalled inside the dome in an upright position, where
on the floor can it be located?
26. Let / be continuous on R = [a, b] x [c, d]. In this
problem, you will establish Leibniz's rule for "differ-
entiating under the integral sign":
d
dy
f
f(x, y)dx
J a
(x, y) dx.
(a) Let G(y') = fy(x, y')dx. For c < y < d, use
the fundamental theorem of calculus to compute
d/dy P G(y')dy' and, therefore,
d
dy
- / / fy{x,y')dxdy' .
y Jc J a
(b) Use Fubini's theorem and part (a) to establish
Leibniz's rule.
27. The function f(x, y) = 1/ ^fxy is unbounded when ei-
ther x or y is zero. Thus, if D = [0, 1] x [0, 1], we
say that j j — dA is an improper double inte-
>xy
gral, analogous to the one-variable improper integral
f1 1
/ — — dx. Improper multiple integrals of this type
Jo V*
may be evaluated using an appropriate limiting pro-
cess. In this problem, you will determine the value of
j j — d A in the following manner:
Id Jxy
(a) For 0 < e < 1/2, 0<<5<l/2, let D,j =
[e, 1 - e] x [8, 1 - 8]. (Note that D€.s c D.)
Calculate
7(6, 8) :
1
dA.
(b) Evaluate lim I(e, 8). You should obtain a fi-
M)-K0,0)
nite value, which may be taken to be the value of
j j — dA, since D( s "fills out" D as (e, 8)
Id -jxy
(0, 0). (We say that in this case the improper inte-
gral converges.)
28. Imitate the techniques of Exercise 27 to determine if
the improper double integral
a
i
dA
[o,i]x[o,i] x + y
converges and, if it does, find its value. (Hint: You will
need to determine lim„_!.o+ u In u.)
29. Imitate the techniques of Exercise 27 to determine if
the improper double integral
I I\o,
dA
/[o,i]x[o,i] y
converges and, if it does, find its value.
30. Calculate / fD In y ' x1 + y2 dA, where D is the unit
disk x2 + y2 < 1 . Note that the integrand is not de-
fined at the origin, so this is an example of an improper
double integral. Nonetheless, you can find its value by
integrating over the annular region
{(x,y) | € <x2 + y2 < 1}
and taking appropriate limits.
31 . Find the value of the improper triple integral
In ^x2 + y2 + z2dV,
where B is the solid ball x2 + y2 + z2 < 1 . (See Exer-
cise 30.)
32. If D is an unbounded region in R2, the integral
ffD f(x, y)dA is another type of improper double
integral, analogous to one-variable improper integrals
such as /fl°° f(x)dx, f(x)dx, or f™^ f(x) dx. In
this problem, you will determine the value of
IL
i
d x2y3
dA,
where D = {(x, y) | x > 1, y > 1} using a limiting
process.
406 Chapter 5 | Multiple Integration
(a) For a > 1, b > 1, let Da j, = [1, a] x [1, b]. Com-
pute
I(a,b)
dA.
(b) Evaluate lim I (a, b). You should obtain a
(a,b)— >(oo,oo)
finite value, which may be taken to be the value
of f f — r — - dA. (In such a case, we say that the
J Jd *2r
improper integral converges.)
33. Let D = l(x, y) | x > 1, y > 1}. For what values of p
and q does the improper integral
1
11
xPyi
dA
converge? For those values of p and q for which the
integral converges, what is the value of the integral?
34. This problem concerns the improper integral
n + x2 + y2y]dA,
R
where p is a constant.
(a) Determine if the integral converges when p = —2
by integrating over the disk Da = {(x, y) \ x2 +
y2 < a2} and then letting a — > oo.
(b) Determine for what values of p the integral
//jj2(1 + x2 + y2)p dA converges. What is the
value of the integral when it converges?
35. Determine if
SSL
1
R3 (1 + x2 + y2 + z2fl2
dV
converges by integrating over the ball Ba = [(x, y, z)
x2 + y2 + z2 < a2} and then letting a — > oo.
36. Show that
SSL
-Jx^+Z1 dy
converges by determining its value.
37. In this problem, you will find the value of the one-
variable improper integral f™ e~x dx by using two-
variable improper integrals.
(a) First argue that J°° e~xl dx converges. (Hint:
Note that e~xl < l/x2 for all x; compare
integrals.)
(b) Let I denote the value of e~xl dx. Show that
/OO rC
-oo J — c
e x y dx dy
/r2
(c) Let Da denote the disk x2 + y2 < a2. Evaluate
ffDae-*2-y>dA.
SL<-'
dA.
(d) Compute I2 as lima^oo
(e) Now find e~xl dx.
'Da
dA.
Exercises 38-46 involve the notion of probability densities. A
probability density function of a single variable is any func-
tion f(x)such that f(x) > 0 for all x G R, and f(x) = 1.
Given such a density function, the probability that a randomly
selected number x falls between the values a and b is
Prob(a < x < b)
f
J a
f{x)dx.
38. (a) Check that f(x) = e 2'*' is a probability density
function.
(b) Egbert turns on the stove in a random manner to
heat cooking oil to fry chicken. If the probability
density of the temperature x of the oil is given by
f(x) = ie-lx-30°l, what is the probability that the
oil has a temperature between 250 °F and 350 °F?
A joint probability density function for two random variables
x and y is a function fix, y) such that
(i) f(x, y) > 0 for all (x, y) e R2, and
(ii) //R2 /(*, y) dA = JZo f-oo y)dx dy=\.
If f is such a probability density and D is a region in R2, then
the probability that a randomly chosen point (x, y) lies in D is
Prob((jc, y) e D) =
39. (a) Show that the function
I2x + y
140
0
f(x,y)dA.
if 0 < x < 5, 0 < v < 4
otherwise
is a joint probability density function,
(b) Find the probability that x < 1 and y < 1 .
40. (a) Show that the function
f(^y)
ye
0
if x > 0, y > 0
otherwise
is a joint probability density function,
(b) What is the probability that x + y < 2?
41 . If a and b are fixed positive constants, what value of C
will make the function f(x, y) = Ce~a^~h^ a joint
probability density function?
42. Let a and b be fixed nonnegative constants, not both
zero. For what value of C is
fi^y)
\C(ax + by) if 0 < x < 1, 0 < y < 1
0 otherwise
a joint probability density function?
Miscellaneous Exercises for Chapter 5 407
43. Let a and b be fixed positive constants and, for a given
constant C, consider the function
f(x,y)
Cxy if 0 < x < a, 0 < y <b
0 otherwise
(a) For what value of C is / a joint probability density
function?
(b) Using the value of C that you found in part (a),
what is the probability that bx — ay > 01
44. The research team for Vertigo Amusement Park deter-
mines that the length of time x (in minutes) a customer
spends waiting to participate in the new Drown Town
water ride, and the length of time y actually spent in the
ride, are jointly distributed according to the probability
density function
-j:/50-v/5
fix, y)
250"
0
if* > 0,y > 0
otherwise
Find the probability that a customer spends at most
an hour involved with the ride (both waiting and
participating).
45. Suppose that you randomly shoot arrows at a circular
target so that the distribution of your arrows is given
by the probability density function
fix, y)
1
where x and y are measured in feet. In the center of
the target, there is a bull's-eye that measures 1 ft in
diameter. (See Figure 5.129.) What is the probability
of your hitting the bull's-eye?
Figure 5.129 The circular
target of Exercise 45. The
shaded region is the
bull's-eye.
46. If x is a random variable with probability density func-
tion f(x) and y is a random variable with probability
density function g(y), then we say that x and y are
independent random variables if their joint density
function is the product of their individual density func-
tions, that is, if
F(x,y) = f(x)g(y).
Suppose that an electrical circuit is designed with two
identical components whose time x to failure (mea-
sured in hours) is given by an exponential probability
density function
fix) =
0 if x < 0
_J_g-Jt/2000 jf > 0 '
2000e
Assuming that the components fail independently,
what is the probability that they both fail within 2000
hours?
47. Let IL denote the moment of inertia of a solid W (of
density S(x, y, z)) about the line L. (See formula (8) in
§5.6.) Let L be the line parallel to L that passes through
the center of mass of W. Then, if M denotes the mass
of W and h the distance between L and L, prove the
parallel axis theorem:
48.
49.
If
Mh1
(Hint: Without loss of generality, you can arrange
things so that L is the z-axis.)
Let F be a function of one variable that is continuous on
the interval [a, b]. Then the function f(x, y) = F{x)
(i.e., the function F considered as a function of two
variables) is continuous on the rectangle R = [a, b] x
[c, d].
(a) Show that the trapezoidal rule approximation Tln n
t0 / [r /(*> y) dA is =id ~ c)T>n, where Tm
denotes the trapezoidal rule approximation to the
definite integral F(x)dx.
(b) Similarly, show that the Simpson's rule approxi-
mation S2m.2„ to ffR f(x, y)dA is S2m,2„ = (d -
c)S2m, where S2,„ denotes the Simpson's rule ap-
proximation to the definite integral fb F(x) dx.
Suppose that / is a function of one variable that is
continuous on [a, b], and g is another function of one
variable that is continuous on [c, d]. Then, from Exer-
cise 40 in 85.2, we know that
l l[a
fix)giy)dA
[a,b]x[c,d]
f{x)dx\U giy)dy
Show that this same product property holds for
the trapezoidal rule approximation. That is, if
Tmi„ denotes the trapezoidal rule approximation
t0 IL
[a,b]x[c,d]
f(x)g(y)d A, then
T,nif)Tn(g),
where Tm(f) denotes the trapezoidal rule approxima-
tion to /* f(x)dx and Tn(g) the trapezoidal rule ap-
proximation to J!! giy)dy.
Line Integrals
6.1 Scalar and Vector Line
Integrals
6.2 Green's Theorem
6.3 Conservative Vector Fields
True/False Exercises for
Chapter 6
Miscellaneous Exercises for
Chapter 6
x(«) x(f \ \S tek = arclength
*• — J 1L — of segment
Figure 6.1 The sum
Yl"k=\ f(x(tt))^sk approximates
the total charge along an idealized
wire described by the path x.
6.1 Scalar and Vector Line Integrals
In this section, we describe two methods of integrating along a curve in space and
explore the meaning and significance of our constructions. The main definitions
are stated first for parametrized paths. Ultimately, we show that the integrals
denned are largely independent of the parametrization, that instead they reflect
essentially only the geometry of the underlying curve.
Scalar Line Integrals
To begin, we find a way to integrate a function (a scalar field) along a path.
Let x: [a, b] — > R3 be a path of class Cl. Let /: X c R3 — > R be a continuous
function whose domain X contains the image of x (so that the composite /(x(f ))
is denned). As has been the case with every other integral, the scalar line integral
is a limit of appropriate Riemann sums.
Let
a = to < t\ <
< h <
<t„=b
be a partition of [a , b] . Let tk be an arbitrary point in the kth subinterval [f*_i , tk ]
of the partition. Then we consider the sum
(i)
k=l
the
where Ask = ffk^ ||x'(0|| dt is the length of the &th segment of x (i.e.
portion of x denned for tk-\ <t< tk). If we think of the image of the path x
as representing an idealized wire in space and /(x(^)) as the electrical charge
density of the wire at a "test point" x(f^) in the £th segment, then the product
f(x(tk))Ask approximates the charge contributed by the segment of curve, and
the sum in ( 1 ) approximates the total charge of the wire. (See Figure 6. 1 .) To find
the actual charge on the wire, it is reasonable to take a limit as the curve segments
become smaller; that is,
Total charge = lim / f(x(tk ))Ask
all Ait^0 ^
all Alt->0 j'
(2)
since x is of class C .
6.1 | Scalar and Vector Line Integrals 409
The mean value theorem for integrals1 tells us that there is some number f£*
in \tk-i, tk] such that
Ask = f ||x'(?)|| dt = (tk - tk-i) \\x'(tk**)\\ = h'(tk**)\\ Atk-
(Here Atk denotes tk — tk-\.) Since ?,* is an arbitrary point in [tk-i, tk], we may
take it to be equal to . Therefore, by substituting for As* in equation (2), and
letting t*k equal f|*, we have
n
Total charge = lim J] /W/D) Aft
= fbf(x(t))\\x'(t)\\dt.
J a
This last result prompts the following definition:
DEFINITION 1 .1 The scalar line integral of / along the C1 path x is
f /(x(f)) ||x'(?)|| dt.
We denote this integral fxfds.
EXAMPLE 1 Let x: [0, 2jt] -> R3 be the helix x(?) = (cos?, sin?, ?) and let
/(x, v, z) = x y + z. We compute
ffds = J f(x(t))\\x'(t)\\dt.
First,
so that
We also have
x'(t) = (— sin?, cos?, 1),
s'(0|| = v7 sin2 ? + cos2 f + 1 = \/2.
/(x(/)) = cos? sin? + f = i sin 2? + ?
from the double-angle formula. Thus,
z1 /• zjt z1 zjt
fds= {\ sin2? + ?) Vldt = V2 / (| sin2? + ?)
= V2 (- 1 cos 2? + \t>) |f = V2 ((- \ + 2x*) -(4 + 0))
2V27T
2
1 Recall that this theorem says that if F is a continuous function, then there is some number c with
a < c < b such that /* F(x)dx = (b - a)F(c).
410 Chapter 6 | Line Integrals
Figure 6.2 ApiecewiseC1
path x.
Given the discussion preceding the formal definition of the scalar line integral,
it is both convenient and appropriate to view the notation fx f ds as suggesting
that the line integral represents a sum of values of / along x times "infinitesimal"
pieces of arclength of x.
Definition 1 . 1 is made only for paths x in R3 and functions / defined on
domains in R3. Nonetheless, for arbitrary n, we may certainly use the definite
integral
f f(x(t)) \\x'(t)\\ dt,
J a
where x is a C1 path in R" and / is an appropriate function of n variables. We
call this definite integral the scalar line integral as well (and maintain the notation
J f ds) and rely on the context to make clear the dimensionality of the situation.
Also, if x is not of class C1, but only "piecewise C1" (meaning that x can be
broken into a finite number of segments that are individually of class C1), then
we may still define the scalar line integral f f ds by breaking it up in a suitable
manner. A similar technique must be used if / (x(?)) is only piecewise continuous.
EXAMPLE 2 Let f(x, y) = y - x and let x: [0, 3] -> R2 be the planar path
(2t, t) ifO < t < 1
x(/) =
(r+ 1,5-4/) ifl <t <3'
Hence, x is piecewise C1 (see Figure 6.2); the two path segments defined for t in
[0, 1] and for / in [1, 3] are each of class C1. Thus,
L
f ds
I fds+i f ds,
where xj(/) = (It, t) for 0 < t < 1 and x2(0 = (f + 1, 5 - 4f ) for 1 < t < 3.
Note that
|x;(o| = VI
and fx' (Oil = VT7.
Consequently,
/ fds= f /(x(0) ||x'(0|| dt = f (t-2t)-V5dt =
Jm Jo Jo
V5
V5
2
Also,
Lfds=L
/(x2(0) 1 4(0 II dt
/I<(5
17 (At
■4t)-(t+ l)Wl7dt
■12717.
Hence,
J,
fds
V5
2
12V17.
Vector Line Integrals —
Now we see how to integrate a vector field along a path. Again, let x: [a, b] —> R"
be of class C1. (n will be 2 or 3 in the examples that follow.) Let F be a vector
6.1 | Scalar and Vector Line Integrals 41 '
field defined on a subset X of R" such that X contains the image of x. Assume
that F varies continuously along x.
DEFINITION 1 .2 The vector line integral of F along x: [a, b] R", de-
noted /XF -ds, is
f F- ds = I F(x(t))-x'(t)dt.
Jx J a
We caution you to be clear about notation. In the vector line integral f F • ds,
the differential term ds should be thought of as a vector quantity (namely, the
"differential" of position along the path), whereas in the scalar line integral
/x f ds, the differential term ds is a scalar quantity (namely, the differential of
arclength).
EXAMPLE 3 Let F be the radial vector field on R3 given by F = xi + yj + zk
and let x: [0, 1] -> R3 be the path x(t) = (t, 3t2, 2t3). Then x\t) = (1, 6t, 6t2),
and so
j^¥-ds = J F(x(t))-x'(t)dt
= f (ti + 3t2\ + 2t\)-(i + 6t\ + 6t2k)dt
Jo
= / (t + m3 + nt5)dt = (\t2 + \tA + 2tb)\\ = i.
Jo
As with scalar line integrals, we may define fF-ds when x is only a piece-
wise C1 path by breaking up the integral in a suitable manner.
Definition 1.2 is important for the following reason:
Physical interpretation of vector line integrals. Consider F to be a force
field in space. Then fx F • ds may be taken to represent the work done by F
on a particle as the particle moves along the path x.
(straight-line
path)
Figure 6.3
in moving a
in a straight
The work done by F
particle from A to 5
line is F • As.
To justify this interpretation, first recall that, if F is a constant vector field
and x is a straight-line path, then the work done by F in moving a particle from
one point along x to another is given by
Work = F • As,
where As is the displacement vector from the initial to the final position. (See
Figure 6.3.)
In general, the path x need not be straight and the force field F may not
be constant along x. Nonetheless, along a short segment of path, x is nearly
straight and F is roughly constant, by continuity. Partition [a, b] as usual (i.e.,
take a = to < ■ ■ ■ < tk < ■ ■ ■ < t„ = b) and focus on the &th segment.
412 Chapter 6 | Line Integrals
F(x©)
'Ax j.
x(f*-i) x($
Figure 6.4 Approximating the
work along a segment of the
path x.
(See Figure 6.4.) Then
Work done along &th segment F(x(f^ )) • Ax*.
Since
x(r + AO-x(f) Ax
x(f) = lim = hm — ,
Af^O Af A^O Af
we must have, for Af* = h — f^-i ~ 0 and f*_i < f| < 4, that
x (*D % = "T ■
tk — h-\ Af*
Hence,
Total work % £ F(x(fj?)) • Ax* « F(x(rt*)) • x'(%*)Af*.
jfc=i
Therefore, it makes sense to take the limit as all the Af^'s tend to zero and define
the total work by
//
Work = lim V F(x(f)) • x'(tt)Atk
all Art^0£— ^
= f F(x(t))-x'(t)dt
J a
F-ds.
Other Interpretations and Formulations
Suppose x: [a, b] —> R" is a C1 path with x'(f) 7^ 0 for a < t < b. Recall from
§3.2 that we define the unit tangent vector T to x by normalizing the velocity:
At)
T(/) =
\At)\
We may insinuate this tangent vector into the vector line integral as follows: From
Definition 1.2, we have
fv-ds= f F(x(t))-x'(t)dt.
Jx J a
Thus,
f F • ds = jf * F(x(0) • |T^| ||x'(0|| dt
= / (F(x(O)-T(O) h'(t)\\dt
I
L
(F-T)ds.
(3)
Since the dot product F • T is a scalar quantity, we have written the original vector
line integral as a scalar line integral. We also see that fxF-ds represents the
6.1 | Scalar and Vector Line Integrals 41 3
{scalar) line integral of the tangential component ofF along the path — that is,
how much of the vector field the underlying curve actually "sees." (See Figure 6.5.)
An important interpretation of the vector line integral occurs when x is a
closed path (i.e., when x(a) = x(b)). In this circumstance, the quantity fxF -ds
is called the circulation of F along x. To understand this idea better, suppose
that F represents the velocity vector field of a fluid. Consider the amount of fluid
moved tangentially along a small segment of the path x during a brief time interval
At. (We use r to denote time here so as not to conflict with the use of t for the
parameter variable of the path x.) Since F • T gives the tangential component of
F, we have that
Amount of fluid moved « (F(x(O)Ar . T(f)) As, (4)
where As is the length of the segment of the closed path x. (See Figure 6.6.)
Formula (4) is only approximate because the segment of the path need not be
Figure 6.6 The amount of fluid
transported tangentially along a segment
of the closed path x is approximately
(F(x(f))Ar-T(?)) As.
completely straight (so that T(t) may not be a constant vector), and also because
the vector field F need not be constant over the segment of the path. If we divide
the term in (4) by At, then the average rate of transport along the segment
during the time interval At is (F(x(/)) • T(f)) As. If we now partition the closed
path x into finitely many such small segments and sum the contributions of the
form (F(x(f )) • T(f )) As for each segment, then let all the lengths As tend to
zero, we find that the average rate of fluid moved, denoted AL/At, is given
approximately by
Finally, if we let At
dL/dr, to be
/ \W t
Figure 6.5 The vector line integral
f^F-ds equals /x(F • T) ds, the
scalar line integral of the tangential
component of F along the path.
AL f
— « / (F-T)Js.
^ Jx
Ai
0, we may define the (instantaneous) rate of fluid moved,
— = [ (F-F)ds = [ F-ds,
at A Jx
which is what we have called the circulation.
414 Chapter 6 I Line Integrals
F = xi + yj
"\*2 + y2 =
Figure 6.7 At every point along
the circle, F = x i + y j has no
tangential component.
EXAMPLE 4 The circle x2 + y2 = 9 may be parametrized by x = 3 cos t, y =
3 sinf, 0 < t < 2n, Hence, a unit tangent vector is
-3 sinf i + 3 cosf j
T =
sinz t + 9 cos2 /
= — sin t\ + cos t j =
Now consider the radial vector field F = x i + y j on R2. At every point along the
circle we have
-y\ + x\
= 0.
F-T = (xi + yj)
Thus, F is always perpendicular to the curve. Therefore,
f¥-ds= j (F-T)ds = j Ods = 0
and considering F as a force, no work is done. Considering F as a velocity field,
the circulation of F along x is likewise zero. (See Figure 6.7.)
On the other hand if instead F = y i — 2x j, then
■•2 2x2
F.T = G>i-2*j).
-yi + xj
-y
3
This quantity will always be negative on the circle, so we expect the circulation
to be negative. In particular, we have
f¥-ds= f(F-T)ds= f -- +2X ds.
Jx Jx Jx 3
Using the parametrization given above, we have ||x'(f)|| = 3, so that
L
y
2x2
9sin2f + 18 cos2/
• 3dt
-L
= -9 (sin2 t + 2 cos2 t)dt = -9 (1 + cos2 1) dt
Jo Jo
= -9 (1 + |(1 +cos2?)) dt
Jo
= -9(|f + isin2/)|^ = -27tt. ♦
Next, suppose that x(?) = (x(f), y(t), z(t)), a < t < b, is a C1 path and
F(x, y, z) = M(x, y, z)i + N(x, y, z)j + P(x, y, z)k
is a continuous vector field. Then, from Definition 1 .2 of the vector line integral,
we have
f¥-ds= f (M(x,y,z)i + N(x,y,z)] + P(x,y,z)k)
Jx J a
■ (x'(t)i + y'(t)j + z'(t)K)dt
= f [M(x, y, z)x'(t) + N(x, y, z)y'{t) + P(x, y, z)z'(t)] dt.
Recall that the differentials of x, y, and z are
dx = x'(t)dt, dy = y'(t)dt, dz = z\t)dt.
6.1 | Scalar and Vector Line Integrals 41 5
Hence,
¥-ds= / M(x,y,z)dx + N(x,y,z)dy + P(x,y,z)dz.
Jx Jx
The integral
M dx + N dy + P dz
L
is a notational alternative to f F • ds. Indeed, the former is defined by the latter.
The integral
L
M dx + N dy + P dz
is commonly referred to as the differential form of the line integral. (In fact, the
expression M dx + N dy + P dz is itself called a differential form.) It empha-
sizes the component functions of the vector field F and arises regularly in applica-
tions. Be sure to interpret M dx + N dy + P dz carefully: It should be evaluated
using the parametric equations for x, y, and z that come from the path x.
EXAMPLE 5 We compute
(y + z) dx + (x + z) dy + (x + y) dz,
where x is the path x(f) = (t, t2, t3) for 0 < t < 1.
Along the path, we have x = t, y — t2, and z = t3 so that dx = dt, dy =
2t dt, and dz = 3>t2 dt. Therefore,
(y + z) dx + (x + z) dy + (x + y) dz
= f (t2 + t3) dt + (t + t3)2t dt + (t + t2)3t2 dt
Jo
L
= / (5t« + 4t* + 3tz)dt = (t:> + t4 + ti)\:. = 3.
Line Integrals Along Curves:
The Effect of Reparametrization
Since the unit tangent vector to a path depends on the geometry of the underlying
curve and not on the particular parametrization, we might expect the line integral
likewise to depend only on the image curve. We shall see precisely to what degree
this observation is true generally for both vector and scalar line integrals.
We begin with an example. Consider the following two paths in the plane:
x: [0, 2n] R2, x(t) — (cost, sinf)
and
y:[0, 7r] ^ R2, y(0 = (cos2*, sin2f).
416 Chapter 6 I Line Integrals
It is not difficult to see that both x and y trace out a circle once in a counterclockwise
sense. In fact, if we let u(t) = 2t, then we see that y(r) = x(m(/)). That is, the path
y is nothing more than the path x together with a change of variables, which
suggests the following general definition:
DEFINITION 1.3 Let x: [a, b] -> R" be a piecewise C1 path. We say that
another C1 path y: [c,d] — > R" is a reparametrization of x if there is
a one-one and onto function u: [c, d] -> [a, b] of class C1, with inverse
u~x : [a, b] [c, d] that is also of class C1, such that y(f) = x(m(?)); that is,
y = x o u.
Reflecting on Definition 1.3 should convince you that any reparametrization
of a path must have the same underlying image curve as the original path.
EXAMPLE 6 The path
x(0 = (1 + It, 2 - t, 3 + 50, 0<f<l,
traces the line segment from the point (1, 2, 3) to the point (3, 1, 8). So does the
path
y(r) = (l + 2?2,2-f2,3 + 5f2), 0 < t < 1.
We have that y is a reparametrization of x via the change of variable u(t) = t2.
However, the path z: [— 1 , 1] -> R3 given by
z(0 = (l+2f2,2-r2,3 + 5f2)
is not a reparametrization of x. We have z(f) = x(u(t)), where u(t) = t2, but in
this case u maps [— 1 , 1] onto [0, 1] in a way that is not one-one.
On the other hand,
w(0 = (3 - It, 1 + 1, 8 - 50, 0 < / < 1,
is a reparametrization of x. We have w(0 = x(l — t), so the function u: [0, 1] — >
[0, 1] given by u(t) = 1 — t provides the change of variable for the reparametriza-
tion. Geometrically, w traces the line segment between (1, 2, 3) and (3, 1, 8) in
the opposite direction to that of x. ♦
If y: [c, d] — > R" is a reparametrization of x: [a, b] — > R" via the change of
variable u: [c, d] — >• [a, b], then, since u is one-one, onto, and continuous, we
must have either
(i) u(c) = a and u(d) = b, or
(ii) u(c) = b and u(d) = a.
In the first case, we say that y (or u) is orientation-preserving and, in the second
case, that y (or u) is orientation-reversing. The idea is that if y is an orientation-
preserving reparametrization, then y traces out the same image curve in the same
direction that x does, and if y is orientation-reversing, it traces the image in the
opposite direction.
6.1 | Scalar and Vector Line Integrals 417
EXAMPLE 7 Ifx: [a, b] — > R" is any C1 path, then we may define the opposite
path xopp:[a,b] -> R" by
xopP(0 = x(a + b - t).
(See Figure 6.8.) That is, xopp(f) = x(u(t)), where u: [a, b] -> [a, £>] is given by
w(f) = a + b — t. Clearly, then, xopp is an orientation-reversing reparametrization
ofx. ♦
Figure 6.8 A path and its
opposite. jn addition to reversing orientation, a reparametrization of a path can change
the speed. This follows readily from the chain rule: If y = x o u, then
y'(t)=^(x(u(t))) = x'(u(t))u'(t). (4)
So the velocity vector of the reparametrization y is just a scalar multiple (namely,
w'(0) of the velocity vector of x. In particular, we have
Speed ofy= ||y'(0|| = || u'(t) x'("(0) ||
= \u'(t)\ ||x'(m(0)|| = \u'(t)\ - (speed ofx). (5)
Since u is one-one, it follows that either u'(t) > 0 for all t 6 [a, b] or u'(t) < 0
for all t 6 [a, b]. The first case occurs precisely when y is orientation-preserving
and the second when y is orientation-reversing.
How does the line integral of a function or a vector field along a path differ
from the line integral (of the same function or vector field) along a reparametriza-
tion of a path? Not much at all. The precise results are stated in Theorems 1 .4
and 1.5.
THEOREM 1.4 Let x: [a,b] R" be a piecewise C1 path and let f:X C
R" —> R be a continuous function whose domain X contains the image of x.
If y: [c, d] —> R" is any reparametrization of x, then
f fds= [
Jy Jx
fds.
PROOF We will explicitly prove the result in the case where x (and therefore, y)
is of class C1. (When x is only piecewise C1, we can always break up the integral
appropriately.) In the C1 case, we have, by Definition 1 . 1 and the observations in
equation (5), that
f fds= fdf(y(t))\\y(t)\\dt= f f(x(u(t)))\\x'{u(t))\\\u'(t)\dt.
If y is orientation-preserving, then u(c) = a, u(d) = b, and \ur(t)\ = uf(t). Thus,
using substitution of variables,
f(x(u(t)))\\x'(u(t))\\ \u\t)\dt= J f(x(u(t)))\\x'(u(t))\\u'(t)dt
= / f(x(u)) ||x'(w)|| du = / fds.
J a Jx
418 Chapter 6 | Line Integrals
If, on the other hand, y is orientation-reversing, then u(c) = b, u{d) = a, and
\u'(t)\ = —u'(t), since u'(t) < 0. Therefore, in this case,
/d pd
f(x(u(t))) \\x'(u(t))\\ \u'(t)\ dt = f(x(u(t))) ||x'(w(/))|| (-u'(t))dt
= — I f(x(u))\\x'(u)\\du
Jb
— I f(x(u)) ||X'(M)|| du = I fds.
J a Jx
Hence, f f ds = fxf ds in either case, as desired. ■
THEOREM 1.5 Letx: [a, b] -» R" beapiecewiseC1 pathandletF: X C R" -+
R" be a continuous vector field whose domain X contains the image of x. Let
y: [c, d] — > R" be any reparametrization of x. Then
1. If y is orientation-preserving, then f F • ds = fx F • ds.
2. If y is orientation-reversing, then f F • ds = —fx F • ds.
PROOF As in the proof of Theorem 1.4, we consider only the C1 case in detail.
Using Definition 1 .2 for the vector line integral, equation (4), and substitution of
variables, we have
n nd nd
F-ds= ¥(y(t))-y'(t)dt= F(x(u{t)))-x\u(t))u'(t)dt.
Jy J c J c
If y is orientation-preserving, then u(c) = a, u(d) = b, so we have
r d pb p
/ F(\(u(t)))-x'(u(t))u'(t)dt = F(x(u))-x'(u)du = / F- ds.
J c J a Jx
This proves part 1. If y is orientation-reversing, then u(c) = b, u(d) = a, so,
instead, we have
p d pa p
l F(x(m(0)) • x'(m(0) u'(t) dt = F(x(w)) • x'(m) du = - / F • ds,
Jc Jb Jx
which establishes part 2. ■
Simply put, Theorems 1 .4 and 1 .5 tell us that scalar line integrals are entirely
independent of the way we might choose to reparametrize a path. Vector line
integrals are independent of reparametrization up to a sign that depends only on
whether the reparametrization preserves or reverses orientation.
EXAM PLE8 LetF = xi + yj, and consider the following three paths between
(0, 0)and(l, 1):
x(0 = (t,t) 0 < t < 1,
l
2 '
y(0 = (2f,20 0 <r<i,
and
z(0 = (!-*, 1-0 0<?<1.
6.1 | Scalar and Vector Line Integrals 41 9
The three paths are all reparametrizations of one another; x, y, and z all trace the
line segment between (0, 0) and (1, 1) — x and y from (0, 0) to (1, 1) and z from
(1, l)to(0, 0).
We can compare the values of the line integrals of F along these paths:
j^F-ds = F(x(t))-x'(t)dt
= [ (ti + tj)-(i + ])dt = f 2tdt=t2\lQ=U
Jo Jo
[F-ds= [ ' (2M + 2rj)-(2i + 2j)d/ = f
Jy JO JO
((l-r)i + (l-0j)-(-i-j)^
1/2 rl/2 \n
%tdt = 4t2\0' = 1;
lF-ds = l
■ i
\2|
[ 2(t-Y)dt =
Jo
The results of these calculations are just what Theorem 1.5 predicts, since y is
an orientation-preserving reparametrization of x and z is an orientation-reversing
one. ♦
The significance of Theorems 1.4 and 1.5 is not merely that they allow us
occasionally to predict the results of line integral computations but also that they
enable us to define line integrals over curves rather than over parametrized paths.
To be more explicit, we say that a piecewise C1 path x: [a, b] -> R" is closed
if x(a) = x(b). We say that the path x is simple if it has no self-intersections,
that is, if x is one-one on [a, b], except possibly that x(a) may equal x(b). Then,
by a curve C, we now mean the image of a path x: [a, b] -> R" that is one-
one except possibly at finitely many points of [a, b]; the (nearly) one-one path x
will be called a parametrization of C. Additionally, we will refer to the curve
C as being closed or simple if it has a corresponding parametrization that is
closed or simple. (See Figure 6.9.) It is a fact (whose proof we omit) that if x
and y are both parametrizations of the same simple curve C, then they must be
reparametrizations of each other.
Not simple, not closed
Simple, not closed
Not simple, closed
Simple, closed
Figure 6.9 Examples of curves.
420 Chapter 6 | Line Integrals
(0,3)
.J
Figure 6.10 The ellipse
^2/25 + y2/9 = 1 of Example 9
is parametrized in the counter-
clockwise sense by x(f ) =
(5 cost, 3 sinf), 0 < t < lit, and
in the clockwise sense by y(r) =
(5 cos2(;r — t), 3 sin2(7r — t)),
0 < t < jr.
or
CO
or
CO
Figure 6.1 1 The possible
orientations for a nonclosed and
a closed, simple curve.
EXAMPLE 9 The ellipse x2/25 + y2/9 = 1 shown in Fi gure 6. 1 0 is a simple,
closed curve that may be parametrized by either
x: [0, 2jt] -> R2, x(f) = (5cosf, 3 sin?)
or by
y:[0, 7t] -> R2, y(f ) = (5 cos 2(jt - f). 3sin2(jr - /)),
since both paths have the ellipse as image and each map is one-one, except at the
endpoints of their respective domain intervals. Note that y is a reparametrization
of x.
However, the path
z: [0, 6jt] -> R2, z(r) = (5cosf, 3 sinf)
is not a parametrization of the ellipse, since it traces the ellipse three times as t
increases from 0 to 6jt. That is, z is not one-one on (0, 6jt). ♦
Simple curves, whether closed or not, always have two orientations that cor-
respond to the two possible directions of travel along any parametrizing path. (See
Figure 6.1 1.) We say that a simple curve C is oriented if a choice of orientation
is made.
The reason for the preceding terminology is that if C is a (piecewise C1)
simple curve, we may unambiguously define the scalar line integral of a continuous
function / over C by
Lfds = l
fds,
where x is any parametrization of C. Theorem 1.4 guarantees that the choice of
how to parametrize C will not matter.
On the other hand, we can define the vector line integral only for oriented
simple curves. If an orientation for C is chosen and x: [a, b] — > R" is a parame-
trization of C that is consistent with this orientation, then we define the vector
line integral of a continuous vector field F over C by
F • ds.
Theorem 1.5 shows that this definition is independent of the choice of para-
metrization of C, as long as it is made consistently with the given orienta-
tion. Indeed if C denotes the same curve as C, only oriented in the opposite
way, then C is parametrized by xopp (where x parametrizes C) and we have,
by Theorem 1.5,
I F-ds= [ F-ds = - J
F-Js
-I
F-ds.
EXAMPLE 10 Let C be the upper semicircle of radius 2, centered at (0, 0)
and oriented counterclockwise from (2, 0) to (—2, 0). Then we may calculate
Jc(x2 — y2 + 1) ds by choosing any parametrization for C. For instance, we may
parametrize C by
x(t) = (2 cos?, 2 sinf), 0 < t < it
6.1 | Scalar and Vector Line Integrals 421
J (x2-y2+l)d
or by
y(t) = (-2cos2f, -2sin2f), -7t/2 <t<0.
(Note that y(t) = x(2t + n).) Then
j (x2 - y2 + l)ds = j{x2 -y2 + l)ds
= / (4 cos2 t - 4 sin2 t + 1)V 4 sin2 1 + 4 cos2 1 dt
Jo
= f (4cos2/ + 1)2 dt = 2(2sin2? + t)\n = 2it,
Jo
by the double-angle formula. Similarly, we check
= / (4 cos2 2t - 4 sin2 2r + 1)V 16 sin2 2? + 16 cos2 2t dt
J-n/2
/o
(4cos4f + 1) • 4dt = 4(sin4f + 0|° /2 = 2jr-
-jr/2 "
(2, 12, 0) EXAMPLE 11 We calculate the work done by the force
F = xi - yj + (x + y + z)k
Figure 6.1 2 The oriented curve on a particle that moves along the parabola y = 3x2, z = 0 from the origin to the
of Example 11. point (2, 12, 0). (See Figure 6.12.)
We parametrize the parabola by x = t, y = 3t2, z = 0 for 0 < t < 2. Then,
by Definition 1 .2,
Work:
IC Jx JO
= f F-ds= f¥-ds= f F(x(t))-x'(t)dt
Jc Jx Jo
= f (t,-3t2,t + 3t2)-(l,6t,0)dt = f (t-m^)dt
Jo Jo
= {\t2-\t% = 2-12=-lQ.
The meaning of the negative sign is that by moving along the curve in the indicated
direction, work is done against the force. If we orient the curve the opposite
way, however, then the work done in moving from (2, 12, 0) to (0, 0, 0) would
be 70. ♦
Numerical Evaluation of Line Integrals (optional)
If we have a scalar line integral fc f ds, or a vector line integral fc¥-ds, and a
suitable parametrization x of C, then Definitions 1 . 1 and 1 .2 enable the evaluation
of the line integrals by means of definite integrals in the parameter variable. These
definite integrals may be difficult — or even impossible — to evaluate exactly, so
422 Chapter 6 | Line Integrals
we might need to resort to a numerical method (such as the trapezoidal rule or
Simpson's rule) to approximate the value of the integral.
Ao 1
Figure 6.1 3 The curve C with n
points chosen on it.
If a (vector) line integral is given in differential form as fc M dx + N dy +
P dz (or as fc M dx + N dy if we are working in R2), then we can give numerical
approximations without resorting to any parametrization as follows. First, choose
points xo, Xi , . . . x„ along C, where xo is the initial point and x„ is the terminal
point. (See Figure 6.13.) For k — 0, . . . , n, let us write
x* = (xk, yic, Zk)
and for k = 1, . . . , n, let
Axk =xk- xk-i, Ayk = yk - yt-u Azk = Zk - Zfc-i-
Finally, let xk = (xk, yk, z*k) denote any point on the arc of C between x^-i and
%k- Then we may approximate the line integral as
/ M dx + N dy + P dz « V [M(x%)Axk + N(x*k)Ayk + P(x*k)Azk] . (6)
Jc k=i
Besides the approximation given in (6), we may also use a version of the
trapezoidal rule adapted to the case of line integrals. In particular, the trapezoidal
rule approximation Tn to the line integral jc M dx + N dy + P dz is
k=l L
(M(x*_o + M(x,)) ^ + (^(^-0 + ^(x*))
+ (P(x,_0+JP(x,))^
fc=l k=\
= (M(xk-0 + M(xk)) ^ + (Mxk-i) + N(xk)) Al
n a
+ J](JP(x,_i) + P(x,))-^. (7)
k=\ 1
6.1 | Scalar and Vector Line Integrals 423
Similar (and shorter) formulas, analogous to (6) and (7), may be given to approx-
imate the two-dimensional line integral fc M dx + N dy;we will not state them
explicitly.
EXAMPLE 12 Let C be the ellipse x2 + 4y2 = 4, oriented counterclockwise.
(See Figure 6.14.) We approximate fc y2 dx + x dy using the (two-dimensional
version of) the trapezoidal rule (7).
,x1
yo = x4
x3
Figure 6.14 The ellipse C in
Example 12.
To make this approximation, let
x0 = (2, 0), xi = (0, 1), x2 = (-2, 0), x3 = (0,-1), x4 = (2, 0).
Then we have
Axi = A^2 = —2, AX3 = AX4 = 2;
A^ = 1, Ay2 = Ay3 = -1, Ay4 = 1.
Hence, from (7), we compute
TA = (02 + l2)f + (l2 + 02)^ + (02 + (-l)2)f + ((-l)2 + 02)§
+ (2 + 0)i + (0 + (-2))^ + ((-2) + 0)=± + (0 + 2)\
= -1-1 + 1 + 1 + 1 + 1 + 1 + 1=4.
The exact answer may be found using the parametrization
Ix = 2cosf
y = sint
We calculate that
0 < t < 2tt.
I y2 dx + x dy = / (sin2 t(— 2 sinf) + 2 cos2 f) dt
Jc Jo
-2tj
f
/ (2(1 - cos2 f)(- sinf) + 1 + cos2f) dt
Jo
= (2
cos t
\ cos3 1 + t + \ sin2r) |^ = 2jt.
From this we see that our approximation is quite rough. It can be improved
by increasing n while taking smaller values of Axk and Ay^. For instance (see
424 Chapter 6 | Line Integrals
Figure 6.15 The ellipse C in Example
12 with eight points marked.
Figure 6.15), if we let
x0 = (2, 0), Xl = (V3, i) , x2 = (0, 1), x3 = (-73, |) , X4 = (-2, 0),
x5 = (-V3, -i) , x6 = (0, -1), x7 = (y/3, -i) , x8 = (2, 0),
then
Axi = V3 - 2, A^2 = Ajf3 = -V?>, AX4 = ^3 - 2, Ax5 = 2 - V^,
Ax6 = Ax7 = V^, Ax8 = 2 - V^;
Ay! = Ay2 = L Ay3 = Ay4 = Ay5 = Ay6 = -\, Ay7 = Ay8 = \.
Therefore, fc y2 dx + x dy is also approximated by
T*=(02 + (tf)^ + ((tf+l2)^ + ((l2+2-)2)
V3
-V3
2 ' " / 2 + Vv"
+ (G)2 + o2)^ + (o2 + (-i)2)
+ ((4)2 + (-D2)f + ((-d2 + (4)2)
+ ((- \f + 02) ^ + (2 + ^ + (V3 + 0)f
+ (0 + (-V3))^ + ((- V3) + (-2))^
+ ((-2) + (- V3))4^ + ((" V3) + 0)4^
2- V3
V3
+ (0 + V3)^ + (V3 + 2)^
= 2 + 2^3 R» 5.4641.
Although still rough, this represents a better approximation.
6.1 | Scalar and Vector Line Integrals 425
EXAMPLE 13 Let C be the line segment from (0, 0, 0) to (2, 2, 2), and con-
sider the line integral fc y V 1 + x 3 dx + e~xz dy + cos (z2) dz. The standard
parametrization of the line segment C by
x = t
y=t 0<t<2
z = t
enables us, in theory, to evaluate the line integral above by evaluating the
one-variable integral
1 + /3 + e~'~ + cos
(t2))
dt.
Unfortunately, we cannot do so exactly (i.e., by means of the fundamental
theorem of calculus, since an antiderivative of the integrand cannot be found
in terms of elementary functions). Instead we provide approximations using
formulas (6) and (7).
Let n = 4. If we choose points along C that are regularly spaced then we
have
x0 = (0, 0, 0), Xl = (i, \,\), x2 = (1,1,1), *3 = (§,§,§), *4 = (2> 2' 2),
so that
Axk = Ayk = Azk = \.
To calculate an approximation using formula (6), we can, for example, let
X^ = l(Xjt_l + xk),
which is the midpoint of the line segment joining xj_i and X&. Then
y* — ( — I I) x* - (I 1 1\ Y* — (— 5 5\ * _ (I 7 7\
Al — U' 4' 4/ ' 2 — U' 4' 4/ ' 3 — U' 4' 4/ ' 4 — U' 4' 4/ '
and so formula (6) tells us that
J y v7! + *3 dx + e~-vz <iy + cos (z2) <iz
+
+
+
i + G)V«-1/16§ +
cos
J_ 1
16 ' 2
cos
_9_ 1
16 ' 2
+ +e-25/16 1 +CQS 25
16 2
yi + dn + e-^i + cos
49 1
16 ' 2
Rs 5.16422.
426 Chapter 6 | Line Integrals
Using the trapezoidal rule formula (7), we may also approximate the line inte-
gral by
t4 = (o + kyf^W) f + (lv/r+W + f
+ (l/TTP + 1^1+ (|)3) f + (l7l + (l)3 + 2^1+2*) f
+ (e° + e->/4)^ + (e-i/4 + ,-1)1/? + (e-i + ,-9/4)1/2
+ (e-9/4 + ^-4)^ + (cos 0 + cos i)^ + (cos ± + cos 1)^
+ (cos 1 + cos + (cos 5 + cos 4)^
% 5.44874.
As was the case in Example 12, because of the small number of sampling
points used, formulas (6) and (7) provide relatively crude approximations. ♦
6.1 Exercises
1. Let fix, y) = x + 2y. Evaluate the scalar line integral
/ / ds over the given path x.
(a) x(?) = (2 - 3t, 4t - 1), 0 < t < 2
(b) x(f) = (cosf, sinf), 0 < t < tt
In Exercises 2-7, calculate J f ds, where f and x are as
indicated.
2. f(x, y, z) = xyz, x(f) = (t, It, 3t), 0 < t < 2
x + z
3. fix, y, z)
y + z
, x(t) = (t, t, f3/2), 1 < t < 3
4. fix, y, z) = 3x + xy + z3, x(f) = (cos4f, sin4f, 3t),
0 < t < 2jt
5. f(x,y,z)
x2 + y
■ , x(f ) = ie2t cos 3t , e1' sin 3t , e2'),
0 < t < 5
6. fix, y,z) = x + y + z,
1(2*, 0, 0) if0<f<l
(2,3f-3,0) if 1 < f < 2
(2, 3, 2t -4) if2<f<3
7. /(A-,y,z) = 2.x-y1/2 + 2z2,
f(f,f2,0) ifO <t < 1
x(0
Exercises 8— 16, find f F • ds, where the vector field F and
the path x are given.
B.F = xi + y) + zK x(t) = (2t + \,t,3t-l), 0<
t < 1
9. F = iy + 2)i + x j, x(f) = (sinf, -cosf), 0<f<
JT/2
10. F = xi + yj, x(f) = (2f + 1, f + 2), 0 < f < 1
11. F = iy - jc)i + xVj,x(f) = it2, f3), -1 < f < 1
12. F = xi + xy] +xyzk, x(f) = (3 cos t, 2 sinf , 5f ),
0 < f < 2tt
13. F = -3yi + Jcj + 3z2k, x(f) = (2f + 1, f2 + f, e'),
0 < f < 1
14. F = jti + yj-zk, x(f ) = (f , 3f2, 2f3), -1 <f < 1
15. F = 3zi + y2 j + 6zk, x(f) = (cos f, sinf , f/3), 0<
f < 4jt
16. F = y cosz i + x sinz j + jcy sinz2 k, x(f) = (f , f2,
?3), 0 < f < 1
1 7. Determine the value of fxx dy — y dx, where x(f ) =
(cos3f, sin3f), 0 < f < n.
(1,1, t-l) ifl<f<3
6.1 | Exercises 427
18. Find the work done by the force field F = 2xi + j
when a particle moves along the pathx(f) = (t, 3t2, 2),
0 < t < 2.
19. If x = (e2fcos3f,e2'sin3/), 0 < t < 2n, find
f x dx + y dy
L(x2 + y2)3/2'
20. Let C be the portion of the curve y = 2^/x between
(1,2) and (9, 6). Find fc 3y ds.
21 . Let F = (x2 + y)i + (y — x)\ and consider the two
paths
x(f) = (t, t2), 0 < t < 1 and
y(f) = (1 -2f,4?2 -At + 1), 0 < t < 5.
(a) Calculate fxF-ds and / F • ds.
(b) By considering the image curves of the paths x and
y, discuss your answers in part (a).
22. Find the work done by the force field F = x2 y i + z j +
(2x — y) k on a particle as the particle moves along a
straight line from (1, 1, 1) to (2, —3, 3).
23. Let F = (2z5 - 3xy) i - x2 j + .y2z k. Calculate the
line integral of F around the perimeter of the
square with vertices (1, 1 , 3), (— 1 , 1,3), (— 1 , — 1,3),
(1,-1,3), oriented counterclockwise about the z-axis.
24. Evaluate fc(x2 — y) dx + (x — y2) dy, where C is the
line segment from (1, 1) to (3, 5).
25. Find fc x2y dx — (x + y) dy, where C is the trapezoid
with vertices (0, 0), (3, 0), (3, 1), and (1, 1), oriented
counterclockwise.
26. Evaluate fc x2y dx — xy dy, where C is the curve with
equation y2 = x3, from (1, —1) to (1, 1).
27. Evaluate fc y dx — x dy, where C is the portion of the
parabola y = x2, from (3, 9) to (0, 0).
28. Evaluate fc(x — y)2 dx + (x + y)2 dy, where C is the
portion of y = \x\, from (—2, 2) to (1, 1).
29. Evaluate fc xy2 dx — xy dy, where C is the semicir-
cular arc from (0, 2) to (0, —2) traveled clockwise.
30. Find the circulation of F = (x2 — y)i + (xy + x)j
along the circle x2 + y2 = 16, oriented counterclock-
wise.
31. Evaluate fc yzdx — xzdy + xy dz, where C is the
line segment from (1, 1, 2) to (5, 3, 1).
32. Calculate fc zdx + x dy + y dz, where C is the curve
obtained by intersecting the surfaces z = x2 and
x2 + y2 = 4 and oriented counterclockwise around the
z-axis (as seen from the positive z-axis).
33. Show that J T • ds equals the length of the path x,
where T denotes the unit tangent vector of the path.
34. Tom Sawyer is whitewashing a picket fence. The
bases of the fenceposts are arranged in the jcy-plane
as the quarter circle x2 + y2 = 25, x, y > 0, and the
height of the fencepost at point {x,y) is given by
h(x, y) = 10 — x — y (units are feet). Use a scalar
line integral to find the area of one side of the fence.
(See Figure 6.16.)
z
y
x2 + y2 = 25
Figure 6.16 The picket fence of
Exercise 34. The base of the fence is the
quarter circle x2 + y2 = 25, x, y > 0.
35. Sisyphus is pushing a boulder up a 100-ft-tall spiral
staircase surrounding a cylindrical castle tower. (See
Figure 6.17.)
(a) Suppose Sisyphus's path is described parametri-
cally as
x(0 = (5 cos 3t, 5 sin 3?, 10?), 0 < t < 10.
Figure 6.17 Sisyphus's path up
the spiral staircase of Exercise 35.
If he exerts a force with a constant magnitude of
50 lb tangent to his path, find the work Sisyphus
428 Chapter 6 | Line Integrals
does in pushing the boulder up to the top of the
tower.
(b) Just as Sisyphus reaches the top of the tower, he
sneezes and the boulder slides all the way to the
bottom. If the boulder weighs 75 lb, how much
work is done by gravity when the boulder reaches
the bottom?
36. A ship is pulled through a 14-ft-long straight channel
by a line that passes from the ship around a pulley.
Assume that a coordinate system is set up so that the
pulley is at the point (24, 32), and the ship is pulled
along the y-axis from the origin to the point (0, 14). If
the tension on the line is kept at a constant 25 lb, find
the work done in tugging the ship through the channel.
37. Suppose C is the curve y = f(x), oriented from
(a, f(a)) to (b, f(b)) where a < b and where / is
positive and continuous on [a, b]. If F = y i, show that
the value of Jc F • ds is the area under the graph of /
between x = a and x = b.
38. Let F be the radial vector field F = xi+yj + zk.
Show that if x(f), a < t < b, is any path that lies
on the sphere x2 + y2 + z2 = c2, then f F • ds = 0.
(Hint:Ifx(r) = (x(t), y(t), z(t)) lies on the sphere, then
[x(t)]2 + [y(f)]2 + [z(t)f = c2. Differentiate this last
equation with respect to t.)
39. Let C be a level set of the function f(x, y). Show that
/CV/. </■ = <>.
40. You are traveling through Cleveland, famous for its
lake-effect snow in winter that makes driving quite
treacherous. Suppose that you are currently located
20 miles due east of Cleveland and are attempting
to drive to a point 20 miles due west of Cleveland.
Further suppose that if you are s miles from the center
of Cleveland, where the weather is the worst, you can
drive at a rate of at most v(s) = 2s + 20 miles per
hour.
(a) How long will the trip take if you drive on a
straight-line path directly through Cleveland? (As-
sume that you always drive at the maximum speed
possible.)
(b) How long will the trip take if you avoid the middle
of the city by driving along a semicircular path
with Cleveland at the center? (Again, assume that
you drive at the maximum speed possible.)
(c) Repeat parts (a) and (b), this time using v(s) =
(s2/16) + 25 miles per hour as the maximum
speed that you can drive.
41 . Consider a particle of mass m that carries a charge q .
Suppose that the particle is under the influence of both
an electric field E and a magnetic field B so that the
particle's trajectory is described by the path x(r) for
a < t < b. Then the total force acting on the particle
is given in mks units by the Lorentz force
F = q(E + v X B),
where v denotes the velocity of the trajectory.
(a) Use Newton's second law of motion (i.e., F = ma,
where a denotes the acceleration of the particle)
to show that
ma • v = qTZ • v.
(b) If the particle moves with constant speed, use part
(a) to show that E does no work along the path
of the particle. (Hint: Apply Proposition 1.7 of
Chapter 3 to v.)
42. Let C be the segment of the parabola y = x2 be-
tween (0, 0) and (1,1) and consider the line integral
fc y3 dx — x2 dy.
(a) Let x0 = (0, 0), X] = (±, £), X2 = (I , I), X3 =
(|, yg), X4 = (1, 1). Use the trapezoidal rule for-
mula (7) and these points to approximate the line
integral.
(b) Now calculate the exact value of the line integral
and compare your results.
43. Let C be the line segment from (0, 0, 0) to (1, 2, 3)
and consider the line integral fc yz dx + (x + z)dy +
x2ydz-
(a) Divide the segment into five regularly spaced
points xo, xi , . . . , X4 and use the trapezoidal rule
formula (7) to approximate this line integral.
(b) Compare your approximation with the exact value
of the line integral.
44. Suppose that magnetic field measurements are made
along a wire shaped as a curve C and the results are
tabulated as follows:
k
Point xt =
B(jct, yk, Zk) =
{Xk, yt, Zk)
Mi+ N j + Pk
0
(-1,-2,-1)
k
1
(0,1,-1)
j + 2k
2
(0, 2, 0)
i + j + 2k
3
(1.2, 1)
2i + j + 2k
4
(1,2,2)
2i + 2j + 2k
5
(1,1,2)
2i + 3j + 3k
6
(1,1,1)
3i + 3j + 3k
7
(1,0, 0)
4i + 3j + 3k
8
(0, 0, 0)
4i + 3j + 4k
By writing / B • ds as fc M dx + N dy + P dz, esti-
mate the work done by B along C using
(a) a trapezoidal rule approximation;
(b) a trapezoidal rule approximation using only the
points x0, x2, x4, x6, x8.
6.2 I Green's Theorem 429
6.2 Green's Theorem
Green's theorem relates the vector line integral around a closed curve C in R2
to an appropriate double integral over the plane region D bounded by C. The
fact that there is such an elegant connection between one- and two-dimensional
integrals is at once surprising, satisfying, and powerful. Green's theorem, stated
generally, is as follows:
C= SD
Figure 6.18 The shaded region D
has a boundary consisting of two
simple, closed curves, each of class
C1, whose union we call C.
THEOREM 2.1 (Green's theorem) Let D be a closed bounded region in R2
whose boundary C = 3D consists of finitely many simple, closed piecewise C1
curves. Orient the curves of C so that D is on the left as one traverses C. (See
Figure 6.18.) Let F(jc, y) = M(x, y)i + N(x, y)\ be a vector field of class C1
throughout D. Then
j> Mdx + Ndy = ff (
dN
dx
dM
)dx dy.
(The symbol §c indicates that the line integral is taken over one or more closed
curves.)
(1,1)
Figure 6.1 9 The region
D of Example 1.
EXAMPLE 1 Let F = xy\ + y2] and let D be the first quadrant region bounded
by the line y = x and the parabola y = x2. We verify Green's theorem in this case.
The region D and its boundary are shown in Figure 6.19. 3D is oriented
counterclockwise, the orientation stipulated by the statement of Green's theorem.
To calculate
® F-ds=(f) xydx + y2dy,
JdD JdD
we need to parametrize the two C1 pieces of 3D separately:
\x = t x = 1 - t
Ci : \ . 0<(<1 and C2: I 0 <t < 1.
y = r
y = l-t
(Note the orientations of C\ and C2.) Hence,
<b xy dx + y2dy = J xy dx + y2dy + / xy dx + y2dy
JdD JCi Jc2
= f (t -t2 + t4 -2t) dt+ f
Jo Jo
= f (t3 + 2t5)dt+ f 2(1 -tf(-dt)
Jo Jo
= (^4+^6)i;+(!a-o3)i:
((1 - tf + (1 - tfX-dt)
_ 1 , 2 _ 2
~~ 4 ~r 6 3
J_
12'
430 Chapter 6 | Line Integrals
On the other hand,
9
dx
9
dx dy = I I —x dy dx = I —x(x — x2) dx
J0 Jx2 JO
= [ (x3 -x2)dx = (
Jo
U4_ lx3\|l
I _ I
4 3
J_
12 •
The line integral and the double integral agree, just as Green's theorem says they
must. ♦
Figure 6.20 The disk of
radius a with boundary
oriented so that Green's
theorem applies.
Figure 6.21 The region
inside the ellipse
x2/a2 + y2/b2 = 1.
EXAMPLE 2 Consider^. — y dx + x dy, where C is the circle of radius a (i.e.,
the boundary of the disk D of radius a), oriented counterclockwise as shown in
Figure 6.20. Although we can readily parametrize C and thus evaluate the line
integral, let us employ Green's theorem instead:
<j> —y dx + x dy = j j
9 9
ir(*) - ^-(-y)
ox ay
dx dy
-IL
2 dx dy = 2(area of D) = lit a
The rightmost expression is twice the area of a disk of radius a. In this case, the
double integral is much easier to consider than the line integral. ♦
The use of Green's theorem in Example 2 can be put in a much more general
setting: Indeed, if D is any region to which Green's theorem can be applied, then,
orienting 3D appropriately, we have
-y dx + x dy
3D
■»//.
2 dx dy = area of D.
(1)
Thus, we can calculate the area of a region (a two-dimensional notion) by using
line integrals (a one-dimensional construction)!
EXAMPLE 3 Using formula (1), we compute the area inside the ellipse
x2/a2 + y2/b2 = 1 (Figure 6.21).
The ellipse itself may be parametrized counterclockwise by
\x = a cos t
0 < t < In.
\y = b sin t
Once again, using formula (1), we find that the area inside the ellipse is
i ® — ydx+xdy=j I —bsint(—asmtdt) + acost(bcostdt)
JdD Jo
-in
= ~ / (ab sin2 1 + ab cos2 1) dt
Jo
ab dt = Ttab.
6.2 I Green's Theorem 431
Alternative Formulations
We rewrite the line integral-double integral formula appearing in the statement
of Theorem 2.1 (Green's theorem) in two ways. These reformulations generalize
to higher dimensions and provide some additional insight in interpreting the
geometric significance of Green's theorem.
To begin, consider a C1 vector field
F = M(x,y)i + N(x,y)j
to be defined on R3 by taking its k-component to be zero. Then, if we compute
the curl of F, we find
\dx~~~dy~)
Therefore, because k • k = 1 , we obtain
Since §SD F • ds = §dD Mdx + Ndy, Green's theorem may be rewritten as
follows:
PROPOSITION 2.2 (A vector reformulation of Green's theorem) If D is
a region to which Green's theorem applies and
F = M(x,y)i + N(x,y)}
is a vector field of class C1 on D, then, orienting 3D appropriately,
* F-ds = f f (Vx F)-kdA.
JBD J J D
To understand this result, visualize the plane region D as sitting in the xy-
plane in R3. (See Figure 6.22.) The vector k is a unit vector normal to D, and
f /D(V x F) • kdA is the double integral of the component of the curl of F normal
to D. Since the line integral §dDF • ds is the circulation of F along 3D (see
§6. 1), the equation of Proposition 2.2 tells us that, under suitable hypotheses, the
circulation of a vector field F along the boundary of a plane region is equal to
the total (or net) "infinitesimal rotation" of F over the entire region. (See also
§3.4 where the curl of a vector field is given an intuitive interpretation in terms of
rotation — or wait until §7.3 when the notion of curl measuring rotation of a vector
field is explained more precisely.) This result generalizes to Stokes's theorem
in R3. Stokes's theorem relates the integral of the component of the curl of a
three-dimensional vector field F that is normal to a surface in R3 to the line
integral of F over the boundary curves of the surface.
Next, we reformulate Green's theorem in another way Once again, assume
that D is a region in R2 to which Green's theorem applies and that its boundary
curves are oriented appropriately. At each point along the C1 segments of 3D,
let n denote the unit vector that is perpendicular to 3D and points away from the
V x F =
l
d/dx
M
J
d/dy
N
k
3/3;
0
432 Chapter 6 | Line Integrals
Figure 6.23 The outward unit
normal n to the region D.
region D. (We call n the outward unit normal vector to D. See Figure 6.23.)
Then we can demonstrate the following:
THEOREM 2.3 (Divergence theorem in the plane) If D is a region to which
Green's theorem applies, n is the outward unit normal vector to D, and
F = M(x, y)\ + N(x, y)j
is a C1 vector field on D, then
(b F-nds =11 V-FdA.
JdD J J D
PROOF Ifx(r) = (x(t), y(t)),a < t < b, parametrizes a C1 segment of 3D, then
along this segment the unit vector n may be obtained geometrically by rotating
the unit tangent vector x'(r)/ 1| x'(t) || clockwise by 90°. In particular, along such a
parametrized C1 segment,
n= /(Qi-x'(f)j =y'(t)i-x'(t)j
Vx'(f)2 + fit)2 uncoil
We calculate the line integral <faD F-nds. Along each C1 segment of 3D,
the contribution to the line integral may be evaluated as
b
(F(x(0) • n(r)) IkCOl dt
= / (M(x(t), y(t))i+N(x(t), y(O)j) ■ , v.„ J ||x'(0|| dt
J a x (')
= f (M(x(t), y(t))y'(t) - N(x(t), y(t))x'(t)) dt
= I -Ndx + Mdy.
J X
Thus, by Green's theorem,
& F-nds=i -Ndx + Mdy =11 [ — - d{-~N^ ) dA
JdD JdD J Jd\3x dy J
= 11
-11
dM dN\
dx dy J
\dA
V-FdA,
D
by the definition of the divergence of F. ■
If C is a simple, oriented curve, the line integral fc F • nds, where n is the
unit normal to C as defined in Theorem 2.3, is known as the flux of F across C.
For example, if F represents the velocity vector field of a planar fluid, then the
flux measures the rate of fluid transported across C per unit time. (We assume
that F does not vary with time t.)
6.2 I Green's Theorem 433
To see this, consider the amount of fluid transported across a small segment
of C during a brief time interval At. As suggested by Figure 6.24, we have
where As is the length of the segment of the curve C. Formula (2) is only ap-
proximate, because the segment of curve need not be completely straight (so the
parallelogram geometry is only approximate), and because the vector field F need
not be constant (so the term F(x , y) At only approximates a flow line of F over the
time interval). If we divide formula (2) by Af, then the average rate of transport
across the segment during the time interval Af is (F(x , y) • n) As. If we now break
up C into finitely many such small segments, sum the contributions of the form
(F(x , y) • n) As for each segment, and let all the lengths As tend to zero, we find
that the average rate of fluid transport, denoted A M / A t , is given approximately by
Finally, letting Af — »• 0, we define the (instantaneous) rate of transport dM/dt
to be
which is the flux.
In view of the remarks above, Theorem 2.3 tells us that the flux of F across
the boundary of plane region D (i.e., what of F flows across dD) is equal to
the total (or net) divergence of F over all of D. We revisit the notion of flux in
Chapter 7 in the case of a three-dimensional vector field F. In that setting, we are
interested in the flux of F across a surface in R3 ; defining such a concept requires
a surface integral. Then Theorem 2.3 generalizes to three dimensions as Gauss's
theorem (also called the divergence theorem). Gauss's theorem relates the flux
of a three-dimensional vector field F across a closed surface to the triple integral
of the divergence of F over the solid region enclosed by the surface.
Proof of Green's Theorem
We establish Green's theorem (Theorem 2.1) in three major steps. The first two
steps consist of proofs of special cases of Theorem 2. 1 . The third step is an outline
of how the special cases may be used to provide a full proof of the general case.
As a result, we fall short of a complete, rigorous proof of the very general version
of Green's theorem stated in Theorem 2.1. However, what we do prove makes use
of the important geometric ideas of multiple integration and line integrals, and
what we do not prove is rather technical.
Step 1. We establish Green's theorem when D is an elementary region in R2
of type 3. Thus, D can be described in two ways (see Figure 6.25):
D = [(x, y) 6 R2 | y(x) < y < S(x), a < x < b]
= {(x, y) 6 R2 | a(y) < x < P(y), c<y <d}.
Amount of fluid transported % area of parallelogram
« (F(x, y)At -n)As,
(2)
The first description of D is as a type 1 elementary region; the second is as a type
2 region. (Recall that a type 3 region is one that is of both type 1 and type 2.) We
assume that the functions a, f),y, and S are all continuous and piecewise Cl.
434 Chapter 6 | Line Integrals
y = 8(x)
Figure 6.25 A type 3 elementary region D analyzed as both a type 1 and
type 2 region. Note the orientations of the boundary curves.
Viewing D as a type 1 elementary region, we evaluate part of <faD Mdx +
Ndy, namely, §aD Mdx. Note that 3D consists of a lower curve C\ and an upper
curve Cj- If we parametrize these curves as follows:
x = t
C\ : \ > , a < t < b and C2 .
}y = y(t) ~ ~ \y = S(t)
x = t
a < t <
then C2 is oriented opposite to the desired orientation shown in Figure 6.25.
Bearing this in mind we compute
M(x,y)dx = I M{x,y)dx — I M(x,y)dx.
3D JCi Jc2
(Note the minus sign!) Then
(b M(x,y)dx= M(t,y(t))dt- M(t,S(t))dt
JdD J a J a
-f
J a
[M(t,y(t))~ M(t,S(t))] dt.
Now we compare the calculation of §dDMdx with that of / fD—(dM/dy) dA.
We have
b rS(x)
J JD dy J a Jy
dM
~dy~
dy dx
-i
-i
[-M(x, 8(x)) + M(x, y(x))]dx
[M(x, y{x)) - M(x, 8(x))] dx.
(Note that the fundamental theorem of calculus was used here.) Thus, we see that,
in this case,
Mdx
3D
dM
~dy~
dA.
6.2 I Green's Theorem 435
Figure 6.26 The region
fl = fliUD2Ufl3U D4, where
each subregion D,-, i = 1, 2, 3, 4, is
elementary of type 3.
Not part
of dD
Part of dD
\
Part of dD
Figure 6.27 The common
boundary of subregions D; and Dj
is oriented one way as part of 9 D,
and the opposite way as part of
dDj. Hence, fno, Mdx + Ndy +
(fgD Mdx + Ndy will cancel over
this common boundary.
In an analogous manner, we can show
f f dN
dA
dD
by viewing D as a type 2 elementary region. We omit the details, except to say
that both the line integral and the double integral can be shown to be equal to
[N(p(y),y)-N(a(y), y)]dy.
Finally, since D is simultaneously of type 1 and type 2,
M(x, y)dx + N(x, y)dy = (b Mdx
3D J3D
dM
Ndy
dD
Wd 3v
-//.
dA +
IL
dN
dx
dA
dN dM\
dA.
dx dy )
Step 2. Now suppose that D is not an elementary region of type 3, but
that it can be subdivided into finitely many type 3 regions D\, D2, . . . , Dn in
such a way that these subregions overlap at most two at a time and only along
common boundaries. Such a region D would look something like the one shown in
Figure 6.26. By Step 1 , Green's theorem holds for each subregion. Hence, we have
fL(N-
My)dA= I J (Nx — My)dA + / / (Nx-My)dA
J J Di J J D2
J j^{Nx-My)dA
+ ••• +
JdDi
M dx + N dy
M dx + N dy
i
J 3D,,
H h <b Mdx + Ndy.
(3)
At this point, it is tempting to conclude immediately that the sum of the line
integrals in equation (3) is <f8D Mdx + Ndy. However, 3D, may contain more
than only portions of dD. The trick is to note that any portion of 3D, that is not
part of 3D is part of exactly one other 3D7 . Moreover, this overlapping portion
is given one orientation by 3D, and the opposite orientation by 3D7-. When we
take the sum of the line integrals in equation (3), any contributions arising from
the components of the 3 D,- 's that are in the interior of D will cancel in pairs.
(See Figure 6.27.) Therefore,
//.<*
My)dA=(b Mdx + Ndy+i) Mdx + Ndy
JdDi J3D2
H h <j> Mdx + Ndy
J 3D,,
J 3D
Mdx + Ndy;
Green's theorem is established in this case.
436 Chapter 6 | Line Integrals
6.2 Exercises
Step 3. Unfortunately, not all regions described in the statement of Theo-
rem 2.1 can be subdivided into finitely many elementary regions of type 3. Here
is an outline of what we might do to prove Green's theorem in such generality.
First, we claim (without proof) that for regions D described in the statement
of Theorem 2. 1 we can produce a sequence of regions D\ , D2, . . . , Dn , . . . whose
"limit" as n — > oo is D and such that each Dn can be subdivided into finitely many
type 3 elementary regions. Next, we claim that 3D„ -> 3D as n -> oo. Finally,
we need to prove that, as n — >• oo,
j jD (Nx - My) dA -> J JjNx - My) dA
and
Mdx + Ndy <b Mdx + Ndy.
dD„ Jao
Since Green's theorem holds for each Dn (by Steps 1 and 2), we are done.
Historical Note3
The idea that the line integral of a vector field along a closed curve can be related
to a double integral over the region bounded by the curve is frequently attributed
to George Green (1793-1841), a self-educated English mathematician. The result
we have been calling Green's theorem had its origins in a rather obscure 1828
pamphlet published by Green, in which he sought to lay a rigorous mathematical
foundation for the physics of electricity and magnetism. Green's ideas arose from
work in partial differential equations concerning gravitational potentials. Green's
pamphlet subsequently came to the attention of Lord Kelvin (1824-1907), who
had it republished so that, fortunately, Green's results received greater recognition.
Coincidentally, a result similar to Green's theorem was established indepen-
dently (and also in 1828!) by the Russian mathematician Mikhail Ostrogradsky
(1801-1861). Ostrogradsky 's name is sometimes associated to what we call
Green's theorem.
In Exercises 1—6, verify Green s theorem for the given vector 2. F = (x2 — y) i + (x + y2)}, D is the rectangle
field bounded by x = 0, x = 2, y = 0, and y = 1 .
F = M(x,y)i + N(x,y)) 3. F = y i + x2 j, D is the square with vertices (1, 1),
(-1, 1), (-1,-1), and (1,-1).
and region D by calculating both
4. F = 2y i + x j, D is the semicircular region x2 + y2 <
(j) Mdx + Ndy and (Nx - My)dA. a2,y>0.
JdD J JD ,
5. F = 1y'\ — Ax j, D is the elliptical region x +
1 . F = -x2y i + xy2 j, D is the disk x2 + y2 < 4. 2y2 < 4.
2 For details of the type of limit argument we have in mind, see O. D. Kellogg, Foundations of Potential
Theory (Springer, Berlin, 1929; reprinted by Dover Publications, New York, 1954), pp. 113-1 19, where
a limit argument is given in the case of Gauss's theorem, which we explore in §7.3. For a proof of Green's
theorem that avoids the limit argument, see D. V Widder, Advanced Calculus, 2nd ed., (Prentice-Hall,
Englewood Cliffs, 1961; reprinted by Dover Publications, New York, 1989), pp. 223-225.
3 See also M. Kline, Mathematical Thought from Ancient to Modern Times (Oxford Press, New York,
1972), p. 683.
6.2 | Exercises 437
6. F = (x2y + x)i + (y3 — xy2)j, D is the region in-
side the circle x2 + y2 = 9 and outside the circle
x2 + y2 = 4.
7. (a) Use Green's theorem to calculate the line integral
y2dx + x2dy,
where C is the path formed by the square with
vertices (0, 0), (1, 0), (0, 1), and (1, 1), oriented
counterclockwise .
(b) Verify your answer for part (a) by calculating the
line integral directly.
8. Let F = 2>xy i + 2x2 j and suppose C is the oriented
curve shown in Figure 6.28. Evaluate
F • ds
both directly and also by means of Green's theorem.
(2,-2)
Figure 6.28 The oriented curve C of
Exercise 8 consists of three sides of a
square plus a semicircular arc.
9. Evaluate
(x2-y2)dx + (x2 + y2)dy,
where C is the boundary of the square with vertices
(0, 0),(1,0), (0, 1), and(l, 1), oriented clockwise. Use
whatever method of evaluation seems appropriate.
1 0. Use Green's theorem to find the work done by the vec-
tor field
F = (4v-3x)i + (x-4y)j
on a particle as the particle moves counterclockwise
once around the ellipse x2 + Ay2 = 4.
11. Verify that the area of the rectangle R = [0,a] x [0, b]
is ab, by calculating an appropriate line integral.
12. Let a be a positive constant. Use Green's theorem to
calculate the area under one arch of the cycloid
x = a(t — s'mt),
a{\ — cost).
13. Evaluate <fc(x4y5 — 2y)dx + (3x + x5y4)dy, where
C is the oriented curve pictured in Figure 6.29.
y
4
Figure 6.29 The oriented curve C of
Exercise 13.
14. Use Green's theorem to find the area enclosed by the
hypocycloid
x(f) = (a cos3 t, a sin3 t), 0 < t < 2n.
15. (a) Sketch the curve given parametrically by x(f) =
(1 -t2,t3 -t).
(b) Find the area inside the closed loop of the curve.
16. Use Green's theorem to find the area between the
ellipse x2/9 + y2/4 = 1 and the circle x2 + y2 = 25.
17. Show that if D is a region to which Green's theorem
applies, and 3D is oriented so that D is always on the
left as we travel along 3D, then the area of D is given
by either of the following two line integrals:
Area of D
x dy
y dx.
18. Find the area inside the quadrilateral whose vertices
taken counterclockwise are (2, 0), (1, 2), (—1, 1), and
(1,1).
1 9. Suppose that the successive vertices of an n-sided poly-
gon are the points (a\, b\), (02, bi), . . . (an, bn), ar-
ranged counterclockwise around the polygon. Show
that the area inside the polygon is given by
oi b\
«2 bj
+
a2 b2
03 £>3
+ ••• +
«n-l bn-1
a„ b„
+
a„ b„
A] bi
20. Let a be a positive integer throughout this problem.
An epicycloid is the path produced by a marked point
on a circle of unit radius that rolls, without slipping,
438 Chapter 6 | Line Integrals
on the outside of a fixed circle of radius a. If the cen-
ter of the fixed circle is at the origin and the marked
point is at (a, 0) when t = 0, then the epicycloid is
given by the path \(t) = ((a + 1) cos t — cos (a + 1 )f ,
(a + l)sin? — sin(a + l)r). (See Exercise 35 of the
Miscellaneous Exercises for Chapter 1.)
(a) Show that the epicycloid path meets the fixed cir-
cle exactly when t = Inn /a, where n is an integer.
(Hint: This must happen when \\x(t)\\ = a.) Graph
the epicycloid when a = 5,6.
(b) Use an appropriate line integral to find the area
enclosed by the epicycloid.
(c) As the integer a gets larger, what happens to the
ratio of the area calculated in part (b) to that of the
fixed circle?
21. Evaluate the line integral §c 5y dx — 3x dy, where C
is the cardioid with polar coordinate equation r =
1 — sin 6, oriented counterclockwise.
22. (a) Suppose that C is a simple, closed curve that does
not enclose the origin. Use Green's theorem to de-
termine the value of
x dx + y dy
ic xz + r
(b) Now suppose that C is a simple, closed curve that
does enclose the origin. Can you use Green's the-
orem to determine the value of
x dx + y dy
x2 + y2
Explain.
(c) Let C\ and C2 be two simple, closed curves that
both enclose the origin, are both oriented coun-
terclockwise, and do not touch or intersect. Show
that
x dx + y dy f x dx + y dy
X1- + yl JCl x^ + y^
(d) Use the result of part (c) to determine the value of
x dx + y dy
ic *2 + r
where C is a simple, closed curve that encloses the
origin.
23. (a) Use the divergence theorem (Theorem 2.3) to show
that <fc F • rids = 0, where F = 2y i — 3x j and C
is the circle x2 + y2 = I.
(b) Now show (fc F • n ds = 0 by direct computation
of the line integral.
24. Let F = M(x, y)\ + N(x, y) j. The divergence theo-
rem shows that the flux of F across a closed curve C
(i.e., <fc F -nds) is equal to / fD(Mx + Ny) dA, where
D is the region bounded by C. Use Green's theorem
to establish a similar result involving (fc F • Tds, the
circulation of F along C. (See also §6.1.)
25. Let C be any simple, closed curve in the plane. Show
that
26. Show that
3x y dx + x dy = 0.
-y3 dx + (jc3 + 2x + y)dy
is positive for any closed curve C to which Green's
theorem applies.
27. Show that if C is the boundary of any rectangular re-
gion in R2, then
3y) dx + x3y2 dy
depends only on the area of the rectangle, not on its
placement in R2 .
28. Let r = x i + y j be the position vector of any point
in the plane. Show that the flux of F = r across any
simple closed curve C in R2 is twice the area inside C.
29. Let D be a region to which Green's theorem applies
and suppose that u(x, y) and v(x, y) are two functions
of class C2 whose domains include D. Show that
3(m, v)
dA
(mVd) • ds,
where C = 3D is oriented as in Green's theorem.
30. Let f(x, y) be a function of class C2 such that
a2/ a2/
dx2 9v2
0
(i.e., / is harmonic). Show that if C is any closed curve
to which Green's theorem applies, then
9/
By
dx
dx
dy = 0.
31. Let D be a region to which Green's theorem applies
and n the outward unit normal vector to D. Suppose
f(x, y) is a function of class C2. Show that
3/
f f V2fdA=(f
J J D JSL
9n
ds,
where V2/ denotes the Laplacian of / (namely,
V2/ = d2f/dx2 + 92//3y2) and df/dn denotes
V/ • n. (See the proof of Theorem 2.3 for more in-
formation about n.)
6.3 | Conservative Vector Fields 439
6.3 Conservative Vector Fields
Figure 6.30 If F has path-
independent line integrals, then
fc F • ds = fc F • ds for any two
piecewise C1 oriented curves from
A to B.
As seen in §6. 1 , line integrals of a given vector field depend only on the underlying
curve and its orientation, not on the particular parametrization of the curve. In
some special instances, however, even the curve itself doesn't matter, only the
initial and terminal points. A vector field having the property that line integrals of
it depend only on the initial and terminal points of the oriented curve over which
the line integral is taken is said to have path-independent line integrals. We next
state a more careful definition, characterize such vector fields, and explore their
significance.
Path Independence
DEFINITION 3.1 A continuous vector field F has path-independent line
integrals if
/ F-ds = I F-ds
JCi Jc2
for any two simple, piecewise C\ oriented curves lying in the domain of F
and having the same initial and terminal points. (See Figure 6.30.)
(1,1)
Figure 6.31 The curves
Ci and Ci of Example 1.
EXAMPLE 1 Let F = y i — x j and consider the following two curves in R2
from the origin to (1, 1): C\, the line segment from (0, 0) to (1, 1), and C%, the
portion of the parabola y = x2 (the curves are shown in Figure 6.31). These curves
may be parametrized as
: \X 1 0 < t < 1 and C2
[y = t
Then we calculate
x = t
y = t2
o < t < 1.
f F-ds= I
JC\ Jo
(H-fj)-(i + j)df
Jo
Odt = 0,
while
f F-ds= f (/2i-/j).(i + 2fj
Jc2 Jo
i)dt
f
Jo
(t2-2t2)dt =
3' lo
We see that
/ F-ds^ j F-ds,
JCi Jc2
and so F does not have path-independent line integrals.
EXAMPLE 2 Let F = xi + yj. This vector field has path-independent line
integrals — we will see why presently. For the moment, however, we illustrate (not
prove) this fact by considering the parabolic path x: [0, 1] -> R2, x(f) = (t, t2),
as well as the path y: [0,2] -> R2 made up of the two straight segments
y,:[0, 1]^R2, yi(0 = (0,0 and y2: [1, 2] -> R2, y2(r) = (t - 1, 1).
440 Chapter 6 | Line Integrals
Figure 6.32 The paths x
and y (consisting of the
paths yi and y2) of
Example 2.
Both x and y are paths from (0, 0) to (1 , 1) and are shown in Figure 6.32. We have
Jv-ds = j (t\ + t2])-(\ + 2t\)dt
= f (t + 2t3)dt= {\t2 + \tA)\[= 1,
Jo
and
F-ds = / F-ds + / F-ds
Jy Jyi Jy2
j fj'j* + ^ (ft - 1)1 + JM*
(t - \)dt
2' lo ^ 2^' L> \\ 2 1 2
1.
To establish that F has path-independent line integrals, we would need to check
that the value of the line integral of F along any choice of path between any two
points is the same as any other — a prohibitive task. ♦
The following result is a reformulation of the path-independence property:
THEOREM 3.2 Let F be a continuous vector field. Then F has path-independent
line integrals if and only if §CF -ds = 0 for all piecewise C1, simple, closed
curves C in the domain of F.
PROOF First, assume that F satisfies the path-independence property. Suppose
C is parametrized by x(t), a < t < b, where x(a) = x(b) = A. Let B be another
point on C, and break C into two oriented curves C\ and C2 from A to B. One
of these curves — say, C\ — will be oriented the same way as C and the other the
opposite way. (See Figure 6.33.) Thus,
Figure 6.33 The simple, (f) F • ds = f F • ds - f F-Js = 0,
closed curve C consists of Jc JC{ JCl
two oriented curves from A
t0 5 since F has path-independent line integrals.
Conversely, suppose that all line integrals of F around simple, closed curves
are zero. Then given two piecewise C\ oriented, simple curves C\ and C2 with
the same initial and terminal points, let C be the closed curve consisting of C\
and — C2 (i.e., C2 with its direction reversed). Then, we have
(t)F-ds= F-ds+ F-ds= F-ds- F-ds.
JC JCi J-C2 JCi Jc2
If C happens to be simple, then j>c F - ds = 0 by assumption, so
/ F-ds= F-ds,
Jc, JCi
6.3 | Conservative Vector Fields 441
as desired. However, C need not be simple even if C\ and C2 are. (See Figure 6.34.)
If it is possible to break C\ and C2 into finitely many segments so that a segment
C[ of C\ and a segment C'2 of C2 either (i) completely coincide or (ii) together
form a simple, closed curve C, then it is not too difficult to modify the preceding
argument to conclude that
Figure 6.34 The closed curve C
constructed from C\ and the
reverse of C2 need not be simple.
However, it is not always possible to do this, and further technical arguments
(which we omit) are required. ■
We remark that it is not essential to assume that the curves in Definition 3.1
and Theorem 3.2 are simple. We have done so here in order to make the proof of
Theorem 3.5 below more straightforward.
/ F-ds = F • ds.
Jc, Jc2
Gradient Fields and Line Integrals
We describe next a class of vector fields that satisfy the path-independence prop-
erty, namely, gradient fields. Suppose that F is a continuous vector field such that
F = V/, where / is some scalar-valued function of class Cl. (Recall that we
refer to / as a scalar potential of F. We also call F a conservative vector field
as well as a gradient field). Then, along any path x from A = x(a) to B = x(b)
whose image lies in the domain of F, we have
/"f-Js= fvf-ds= f Wf(x(t))-x'(t)dt.
Jx Jx J a
It follows from the chain rule that d/dt[f(x(t))] = V/(x(r)) • x'(0- Hence,
fjf-d* = f V/(x(0) • AO dt = fjt dt
= f(x(t))\ha = f(x(b)) - f(x(a)) = f(B) - f(A).
Therefore, when F is a gradient field the line integral of F depends only on the
value of the potential function at the endpoints of the path. Hence, gradient fields
have path-independent line integrals. The converse holds as well, as we prove at
the end of this section. Stated formally, we have the following theorem:
THEOREM 3.3 Let F be defined and continuous on a connected, open region R
of R". Then F = V/ (where / is a function of class C1 on R) if and only if F has
path-independent line integrals over curves in R. Moreover, if C is any piecewise
C , oriented curve lying in R with initial point A and terminal point B, then
jv.di = f(B)-f(A).
Note: A region R C R" is connected if any two points in R can be joined by
a path whose image lies in R.
EXAMPLE 3 Consider the vector field F = x i + y j of Example 2 again. You
can readily check that F = V/, where f(x, y) = \{x2 + y2). By Theorem 3.3,
line integrals of F will be path independent; this fact was illustrated, but not
proved, in Example 2 when we calculated the vector line integral of F along two
442 Chapter 6 | Line Integrals
paths from (0, 0) to (1, 1). By Theorem 3.3, we see now that for any directed
piecewise C1 curve C from (0, 0) to (1, 1), we have
J^F-ds = f(l, 1)— /(0, 0) = \{\2 + l2) - i(02 + 02) = 1,
which agrees with our earlier computations. ♦
A Criterion for Conservative Vector Fields —
Theorem 3.3 tells us that a vector field F has path-independent line integrals
precisely when it is a conservative (gradient) vector field and moreover, that the
line integral of F along any path is determined by the values of the potential
function / at the endpoints of the path. Two questions arise naturally:
1. How can we determine whether a given vector field F is conservative?
2. Assuming that F is conservative, is there a procedure for finding a scalar
potential function / such that F = V/?
We answer the first question by providing a simple and effective test that can
be performed on F. Should F pass this test (i.e., if F is conservative), then we
illustrate via examples how to produce a scalar potential for F, thereby answering
the second question.
First, we need additional terminology.
DEFINITION 3.4 A region R in R2 or R3 is simply-connected if it con-
sists of a single connected piece and if every simple, closed curve C in R
can be continuously shrunk to a point while remaining in R throughout the
deformation.
If R is a region in the plane, then R is simply-connected just in case it is
connected and every simple, closed curve C lying in R has the property that all
the points enclosed by C also lie in R. Loosely speaking, a simply-connected
region (in either R2 or R3) can have no "essential holes." Illustrative examples are
shown in Figures 6.35 and 6.36. The notion of continuously shrinking a curve to a
point can be made fully precise, although we shall not take the trouble to do so here.
(1) (2)
Figure 6.35 (1) The region Ri C R2 is simply-connected: All points
surrounded by any simple, closed curve in Ri lie in R i . (2) In contrast, R2 is not
simply-connected: Although the curve C\ encloses points that lie in R2, the
curve C2 surrounds a hole. Hence, C2 cannot be continuously shrunk to a point
while remaining in R2 .
6.3 I Conservative Vector Fields 443
(2)
c
{(0,0,0))
R7 = &
■ z-axis
Figure 6.36 (1) The region R\ C R3 is simply-connected. (2) The region R2 is not
simply-connected: The curve C cannot be shrunk continuously to a point without
becoming "stuck" on the "missing" z-axis.
Now we state our criterion for a vector field to be conservative.
THEOREM 3.5 Let F be a vector field of class C1 whose domain is a simply-
connected region R in either R2 or R3. Then F = V/ for some scalar- valued
function / of class C2 on R if and only if V x F = 0 at all points of R.
Before proving Theorem 3.5, some remarks and examples are appropriate.
First, note that Theorem 3.5 provides a straightforward way to determine if a vector
field F is conservative: Check that the domain of F is simply-connected and then
test if V x F = 0. If the curl vanishes, it follows that F has path- independent line
integrals. This "curl criterion" can be helpful in practice.
In the case where F = M(x , y) i + N(x , y) j is a two-dimensional vector field
the condition that the curl of F vanishes means
dN
~dx~
i i k
d/dx d/dy d/dz
M(x,y) N(x,y) 0
This is equivalent to the condition
dN _ dM
dx dy
Equation (1) is a simpler condition to use in this situation.
dM\
k = 0.
dy J
(1)
EXAMPLE 4 LetF = x2yi-2xyj.Then
dx
(— 2xy) = —2y and
dy
(x2y) = x2.
Since these partial derivatives are not equal, we conclude that F is not conservative,
by Theorem 3.5. ♦
EXAMPLE 5 Let F = (2xy + cos 2y)i + (x2 - 2x sin2y)j. The vector field
F is defined and of class C1 on all of R2 (a simply-connected region). Moreover,
— (x — 2x sin2y) = 2x — 2 sin2y and — (2xy + cos2y) = 2x — 2 sin2y.
dx dy
444 Chapter 6 I Line Integrals
We may conclude that F is conservative. In addition, if C is the ellipse
x2/4 + v2 = 1 (a simple, closed curve), then, by Theorems 3.2 and 3.3, we con-
clude, without any explicit calculation, that j>c F • ds = 0. ♦
EXAMPLE 6 Let
F =
x1 + y2 + z2
6x i +
x 2 + yz + z
+ y + Z
F is of class C1 on all of R3 except for the origin. Note that R3 — {(0, 0, 0)} is
simply-connected. We leave it to you to check that V x F = 0 for all (x, y, z) in
the domain of F. Therefore, by Theorem 3.5, F is conservative.
Now suppose x: [0, 1] -» R3 is the path given by x(t) = (1 — t, shur/, t). To
evaluate fx F • ds directly, we must calculate
i
(1 - t)2 + sin2 nt + t2
6(1-0,
sm7rf
(1 -t)2 + sm27tt + t
2 '
(1,0,0)
Figure 6.37 The paths
\(t) = (1 - t, sinjrf, t), 0 < t < 1
and y(f ) = (cos r, 0, sin t ),
0 < ? < jt/2.
(1 - 02 + sin27rf + f:
2r — 1 + it sin itt cositt
(1 - t)2 + sin2 7if + t2
(— 1, 7T COS 7Tf , 1) c/f
+ 6(1 - 01*
This last integral is tricky to evaluate. However, since F is conservative, we may
evaluate F by calculating / F • ds, where y is any other path with the same
endpoints as x. A good choice is y(0 = (cos?, 0, sinO, 0 < t < n/2, because
the image of this path lies on the sphere x2 + y2 + z2 = 1 , a fact that will en-
able us to work with a simple integral. (See Figure 6.37 for a graph of the two
paths x and y.) Since F is conservative (and hence has path-independent line
integrals),
6cosf,
0 sinf
1 1
(— sinf, 0, cos t)dt
-L
n/2
6 cos t sin t dt
-L
n/2
3 s\n2t dt
-|cos2r|0
n/2
K-i-i) = 3.
Sketch of a proof of Theorem 3.5 By Theorem 4.3 of Chapter 3, note that, if
F = V/ for some function / of class C2, then V x F = V x (V/) = 0.
Conversely, suppose that V x F = 0. We show that if C is any piecewise C1,
simple, closed curve in R, then §c F • ds = 0. By Theorem 3.2, this implies that
F has path-independent line integrals, which, by Theorem 3.3, is equivalent to
F's being the gradient of some scalar- valued function /. Moreover, since F is
assumed to be of class C1, it follows that / must be of class C2.
6.3 | Conservative Vector Fields 445
Figure 6.38 Since R is
simply-connected, any region D
enclosed by C must lie in R.
Figure 6.39 A
surface in space,
bounded by the
simple, closed
curve C.
To see that §c^ • ds = 0, suppose, first, that F is defined on a simply-
connected region R in R2. Since R is simply-connected, the closed curve C
bounds a region D that is entirely contained in R. (See Figure 6.38.) By
Proposition 2.2, which is equivalent to Green's theorem, we have
(f ¥-ds = ± f f (VxF-k)JA = ± / /
Jc J Jd J Jd
OdA = 0.
We use the "±" sign in the event that C is oriented opposite to the orientation
stipulated by Green's theorem. If F is defined on a simply-connected region R
in R3, then we must apply Stokes's theorem rather than Green's theorem. This
gets us a little ahead of ourselves, although the principle remains the same: If C
is a simple, closed curve in R C R3, then <fc F • ds is equal to a suitable surface
integral of V x F over a surface in R bounded by C. (See Figure 6.39.) Because
the curl is assumed to be zero, any integral of it will be zero, and so §c F • ds is
zero as well. ■
Finding Scalar Potentials
Now that we have a practical test to determine whether a given vector field F
is conservative, we illustrate in Examples 7 and 8 a straightforward method for
producing a scalar potential function for F. This technique is a direct consequence
of the definition of a gradient field.
EXAMPLE 7 Consider the vector field
F = (2xy + cos 2y) i + (x2 — 2x sin 2 y ) j
of Example 5. We have already seen that F is conservative in Example 5. To find
a scalar potential for F, we seek a suitable function f(x, y) such that
V/(*,y) = — i+ — J = F.
ox ay
The components of the gradient of / must agree with those of F; therefore,
Bf
dx
Bf
3y
We may begin to recover / by integrating the first equation of (2) with respect to
x. Thus,
fdff 7
f(x,y)= I — dx = / (2xy + cos2y)dx = x y + x cos2y + g(y), (3)
J dx J
where g(y) is an arbitrary function of y. (The function g(y) plays the role of the
arbitrary "constant of integration" in the indefinite integral of df/dx.) Differen-
tiating equation (3) with respect to y yields
9/
dy
If we compare equation (4) with the second equation of (2), we see that g'(y) = 0,
and so g must be a constant function. Therefore, our scalar potential must be of
the form
f(x, y) = x2y + x cos2y + C,
= 2xy + cos2y
= x2 — 2x sin2y
(2)
2x sin2y + g'(y).
(4)
446 Chapter 6 | Line Integrals
where C is an arbitrary constant. You may, if you wish, double-check that
V/ = F. ♦
EXAMPLE 8 LetF = (ex siny - yz)i+ (ex cosy - xz)\ + (z - xy)k.Note
that F is of class C1 on all of R3. We calculate
i j k
d/dx d/dy d/dz
ex sin y — yz ex cos y — xz z — xy
3 3
— (z- xy) - — (ex cos y - xz)
ay dz
3 3
+ ( 7^(ex siny - yz) - — (z - xy)
3 3
( g^(ex cosy - xz) - —(ex siny - yz)
= 0.
Therefore, by Theorem 3.5, F is conservative.
Any scalar potential f(x, y, z) for F must satisfy
V
dx
df
8y
df
dz
= e sin y — yz
= ex cos y — xz-
= z — xy
Integrating df/dx with respect to x, we find that
^ dx
= j (ex sin y — yz) dx
= ex siny - xyz + g(y, z),
(5)
(6)
where g(y, z) may be any function of y and z. Differentiating equation (6) with
respect to y and comparing with the second equation in (5), we see that
M x 9s ,
— = e cos v — xz H = e cos y — xz.
dy 3y
Hence, dg/dy = 0, so g must be independent of y; that is, g(y, z) = h(z), a
function of z alone. So
f(x, y,z) = ex siny - xyz + h(z).
(7)
6.3 | Conservative Vector Fields 447
Finally, we differentiate equation (7) with respect to z and compare with the third
equation of (5):
— = -xy + h(z) = z - xy.
dz
Therefore, h'(z) = z, so h(z) = \z2 + C, where C is an arbitrary constant. Thus,
a scalar potential for the original vector field F is given by
f{x, y, z) = ex sin y - xyz + \z2 + C. +
Addendum: Proof of Theorem 3.3
Recall that we have already shown that if a vector field F is a gradient field then F
has path-independent line integrals. So now we need only establish the converse.
We do this explicitly in the case where F is defined on a (connected) subset R of
R3 , although our proof requires only notational modification in the n -dimensional
setting.
Assume that F has path-independent line integrals. Then we may unambigu-
ously write F • ds to denote the vector line integral of F from the point A
to the point B along any path whose image lies in R. In what follows, consider
A(xo, yo, Zo) to be a fixed point in R, and 5(x, y, z) a "variable point." Then we
define
KB)
J A
ds
and show that / is a scalar potential for F.
Write F explicitly as
F = Mix, y,z)i + Nix, y, z)j + P(x, y, z)k.
Therefore, we need to verify that the components of V/ agree with those of F;
that is,
Figure 6.40 If B' is
sufficiently close to B, then the
straight-line path from B to B'
will lie inside R.
df
dx
= Mix, y, z),
df
dy
= Nix, y, z), and
Bf
dz
= Pix,y,z).
Actually, we check only the first of these equations; the others may be verified in
a similar manner.
At the point B, we have, by the definition of the partial derivative, that
3/ ,. fjx + h,y,z)- fjx,y,z) fjB') - fjB)
— = hm = lim ,
dx h-+o h h-+o h
where B' denotes the point (x + h, y, z). Note that B'
merator of the difference quotient in equation (8) is
B as h
fiB')-fiB)= f F-ds- fBF-ds = ^ F
J A J A JB
ds.
(8)
0. The nu-
(9)
If h is sufficiently small, we may evaluate the line integral in equation (9) by using
the straight-line path between B and B' . (See Figure 6.40.) Explicitly, this path is
x(t) = ix + th, y, z), 0 < t < 1.
448 Chapter 6 | Line Integrals
Then
6.3 Exercises
pB' pi pi
/ F-ds = F(x + th, y,z)-(h,0, 0)dt= / hM(x + th, y, z)dt.
Jb Jo Jo
Since t is between 0 and 1, \th\ < \h\. By the continuity of F (and therefore M)
we have, for h s» 0,
M(x + th, y, z) ~ M{x, y, z).
This approximation improves as h — »• 0. Hence,
f(B')-f(B)= f F-ds « / hM(x,y,z)dt = hM(x,y,z).
Jb Jo
Using this result in equation (8), we see that
^ = lim ^-[/iM(x, y, z)] = M(x, y, z),
dx h^o h
as desired.
1. Consider the line integral Jc z2 dx + 2y dy + xz dz.
(a) Evaluate this integral, where C is the line segment
from (0,0, 0) to (1, 1, 1).
(b) Evaluate this integral, where C is the path
from (0,0,0) to (1,1,1) parametrized by
x(f) = (r, ?2,?3), 0 < t < 1.
(c) Is the vector field F = z2i + 2yj+xzk conser-
vative? Why or why not?
2. Let F = 2xy i + (x2 + z2) j + 2yz k.
(a) Calculate fc F • ds, where C is the path param-
etrized by x(f) = (t2, f3, f5), 0 < t < 1.
(b) Calculate Jc F • ds, where C is the straight-line
path from (0, 0, 0) to (1, 0, 0), followed by the
straight-line path from (1, 0, 0) to (1, 1, 1).
(c) Does F have path-independent line integrals? Ex-
plain your answer.
In Exercises 3—17, determine whether the given vector field
F is conservative. If it is, find a scalar potential function for F.
3. F = ex+y i + exy j
4. F = 2x sin y i + x2 cos y j
5. F = (3a-2 cos y + - J—. , ) i
V l+x2y2/
2 2
XV XV
6. F = i H —
(1+x2)2 l+x2'
7. F = (e~y — y sinxy)i — (ic"1' + x sinxy)j
8. F = (6xy2 + 2y3)i + (6x2y-xy)j
9. F = (6xy2 - 3x2)i + (y2 + 6x2y)j
10. F = (xyz3 + xy - z2) i + (2x2z3 - y2 + 2yz) j
+ (6x2y - y2z)k
11. F = (4xyz3 — 2xy) i + (2x2z3 - x2 + 2y z) j
+ (6x2yz2 + y2)k
12. F = (2xz - y2 + yzexvi:)i - (2xy + xzexyz)\
+ (x2 + xyexyz) k
13. F = (2x + y)i + (zcosyz + x)j + (ycosyz)k
14. F = (y + z)i + 2zj + (x + y)k
15. F = ex s'my i + e* cos y j + (3z2 + 2) k
16. F = 3x2i+ — j + 2zlnyk
y
17. F = (e-yz - yzexyz)i + xz(e-yz + exyz)i
+ xy(e~yz - e"z) k
1 8. Of the two vector fields
F = xy2z3 i + 2x2y j + 3x2y2z2 k
and
G = 2xy i + (x2 + 2yz) j + y2 k
one is conservative and one is not. Determine which
is which, and for the conservative field find a scalar
potential function.
19. (a) Let / be a function of class C1 defined on a con-
nected domain in R" . Show that if the gradient of
/ vanishes at all x = (xi, . . . , x„) in its domain,
then / is constant.
6.3 | Exercises 449
(b) Suppose that F is a conservative vector field de-
fined on a connected subset of R" . Show that if g
and h are both class C1 potential functions for F,
then g and h must differ by a constant.
20. Find all functions M(x, y) such that the vector field
F = M(x , y ) i + (x sin y — y cos x) j
is conservative.
21. Find all functions N(x, y ) such that the vector field
F = (ye2* + 3xV)i+ y)j
is conservative.
22. Let G(x, y) = (xc* + y2)i + xy j. Find all functions
g(jc) such that the vector field F = g(x)G is conserva-
tive on all of R2.
23. Find all functions N(x, y , z) such that the vector field
F = (x3y — 3x2z) i + N(x, y , z) j + (2yz - x3) k
is conservative.
24. For what values of the constants a and b will the vector
field
F = (3x2 + 3y2z sinxz)i + (aycosxz + bz)\
+ (3xy2 sinxz + 5y)k
be conservative?
25. Let F = x2 i + cos y sin z j + sin y cos z k.
(a) Show that F is conservative and find a scalar po-
tential function / for F.
(b) Evaluate fxF-ds along the path x: [0, 1] -> R3,
x(t) = (t2 + \,et,e2t).
Show that the line integrals in Exercises 26-28 are path inde-
pendent, and evaluate them along the given oriented curve and
also by means of Theorem 3.3.
26.
L
(3x — 5y) dx + (7y — 5x) dy ; C is the line segment
from (1,3) to (5, 2).
2j f x dx
Jc Jx
x dx + y dy
Jx^+
C is the semicircular arc of x2 +
28
= 4 from (2, 0) to (-2, 0).
. j (2y — 3z) dx + (2x + z) dy + (y — 3x) dz\ C is the
line segment from the point (0, 0, 0) to (0, 1, 1) fol-
lowed by the line segment from the point (0, 1, 1) to
(1,2,3).
In Exercises 29— 32, find the work done by the given vector field
F in moving a particle from the point A to the point B whose
coordinates are as indicated.
29. F = (3x2y - y2)i + (x3 - 2xy)j; A(0, 0), B{2, 1)
30. F = 3^yi + 2x3/2j;A(l,2), S(9, 1)
31. F = (2xyz-y2z3)i + (x2z-2xyz3)j
+ (x2y - 3xy2z2)k; A(l, 1, 1), 5(6, 4, 2)
32. F = 2xy cos z i + x2 cos z j — x2y sin z k;
A(l, l,7r/2), S(2, 3,0)
33. (a) Determine where the vector field
X + xv2 x2 + 1
F = — — 1 —t
r r
is conservative.
(b) Determine a scalar potential for F.
(c) Find the work done by F in moving a particle along
the parabolic curve y = 1 + x — x2 from (0, 1) to
(1,1)-
34. Let /, g, and h be functions of class C1 of a single
variable.
(a) Show that F = (/(x) + y + z) i + (x + g(y) +
z) j + (x + y + h(z)) k is conservative.
(b) Determine a scalar potential for F. (Your answer
will involve integrals of /, g, and h.)
(c) Find fc F • ds, where C is any path from
(x0, y0, zo) to (xi, yi, zi).
35. Consider the vector field
F = (2x + z) cos (x2 + xz) i - (z + 1) sin (y + yz) j
+ (x cos (x2 + xz) — y sin (y + yz)) k.
(a) Determine if F is conservative.
(b) If x(f) = ^3, t2. nt - sin y I, 0<r < 1, evalu-
ate fxF-ds.
36. Consider the vector field
G = (2x + z) cos (x2 + xz) i
+ (x-(z + l)sin(y + yz))j
+ (x cos (x2 + xz) - y sin (y + yz)) k.
(a) How is G different from the vector field F in Ex-
ercise 35? Is G conservative?
(b) If x(f) = ^r3, r2, 7rf - sin ^J, 0 <f < 1, evalu-
ate fxG-ds.
37. Let F be the gravitational force field of a mass M on a
particle of mass m:
7r r
GMm
(x2 + y2 + z2)3/2
(xi + yj + zk).
450 Chapter 6 | Line Integrals
(This is the force field of Example 3 in §3.3.) Given xo = (xq, yo, zo) to Xi = (jci, yi, zi) depends only on
that G, M, and m are all constants, show that the ||xo|| and ||xi||.
work done by F as a particle of mass m moves from
True/False Exercises for Chapter 6
1 . If C is the parabola y = 4 — x2 with — 2 < x < 2, then
Jc y sinx ds = 0.
2. If F = — i + j + k and C is the straight line from the
origin to (2, 2, 2), then f„ F • ds = 2«/3,
3. If F = xi + y] + zk and C is the straight line from
(3, 3, 3) to the origin, then fc F • ds is positive.
4. Suppose that f(x) > 0 for all x. Let F = fix) i. If C
is the horizontal line segment from ( 1 , l)to(2, l),then
fcF-d$> 0.
5. Suppose that f(x) > 0 for all x. Let F = f(x) i. If C
is the vertical line segment from (0, 0) to (0, 3), then
fc¥-ds> 0.
6. If x is a unit-speed path, then f F • ds = f (F • v) ds,
where v denotes the velocity of the path.
7. If x and y are two one-one parametrizations of the
same curve and F is a continuous vector field, then
fxY.ds = fyF.ds.
8. If a nonvanishing, continuous vector field F is every-
where tangent to a smooth curve C, then F does no
work along the curve.
9. If a nonvanishing, continuous vector field F is every-
where normal to a smooth curve C, then F does no
work along the curve.
10. If the curve C is the level set at height c of a func-
tion f(x, y), then fc f(x, y)ds is c times the length
of C.
11. If f(x,y,z) is a continuous function and
Jc f{x, y, z)ds = 0 for all curves C in R3, then
f(x, y, z) = 0 for all (x, y, z) G R3.
12. If a closed curve C is a level set of a function f(x, y)
of class C1 and V/ / 0, then the flux of V/ across C
is always zero.
13. If a closed curve C is a level set of a function f(x, y)
of class C1 and V/ / 0, then the circulation of V/
along C is always zero.
14. If a vector field F has constant magnitude 3 and makes
a constant angle with a curve C, then the work done
by F along C is 3 times the length of C.
15. If F is a continuous vector field everywhere tangent to
an oriented C1 curve C, then Jc F • ds = fc ||F|| ds.
16. IfF is a constant vector field on R2, then <pc F -ds = 0,
where C is any simple, closed curve.
17. IfF is an incompressible (i.e., divergenceless) C vec-
tor field on R2 and C is a simple, closed curve, then
the circulation of F along C is always zero.
18. IfF is an incompressible C1 vector field on R2, then
the flux across any simple, closed curve C in R2 is
always zero.
19. If C is a simple curve in R2, then f„ V/ • ds = 0.
20. If C is a simple, closed curve in R2 and / is of class
C1,thenjfcV/-Js = 0.
21 . F = (ex cos y + 3)i — ex sin y j is a conservative vec-
tor field on R2.
22. If / and g are functions of class C1 defined on a region
D in R2, then
f fVg-ds=£> gVf-ds.
J 3D JdD
23. If C is a closed curve in R3 such that <fc F • ds = 0,
then F is conservative.
24. (fc x dx + y dy + z dz = 0 for all simple, closed
curves C in R3 .
25 . <fc ex (cos y sin z dx + sin y sin z dy + cos y cos z) dz
= 0 for all simple, closed curves C in R3.
26. If V x F = 0, then F is conservative.
27. Let M(x, y) and N{x, y) be C1 functions with domain
R2 - {(0, 0)}. If dM/dy = dN/dx, then §c M dx +
N dy = 0 for any closed curve C in R2.
28. Let M(x,y,z), N(x,y,z), and P(x,y,z) be C1
functions with domain R3 - {(0, 0, 0)}. If dM/dy =
8N/dx, dM/dz = dP/dx, and 8N/dz = dP/dy,
then <fc M dx + N dy + P dz = 0 for any closed
curve C in R3 .
29. If F:R" — > R", then there is at most one function
/: R R such that V/ = F.
30. If F is a differentiable vector field and F = V x G,
then V • F = 0.
Miscellaneous Exercises for Chapter 6 451
Miscellaneous Exercises for Chapter 6
Let C C R" be a piecewise Cl curve and /:XCR" — > R
a continuous function whose domain X includes C. Then we
define the quantity
[/].
fcfds fcfds
, ds length of C
to be the average value of f along C. Exercises 1—5 concern
the notion of average value along a curve.
1 . Explain why it makes sense to use the preceding inte-
gral formula to represent the average value. (A careful
explanation involves the use of Riemann sums.)
2. Suppose that a thin wire is shaped as a helical curve
parametrized by
x(t) = (cost, sin?, t), 0 < t < 3tt.
If fix, y, z) = x2 + y2 + 2z2 + 1 represents the tem-
perature at points along the wire, find the average tem-
perature.
3. Find the average y-coordinate of points on the upper
semicircle y = y/a2 — x2.
4. Find the average z-coordinate of points on the broken-
line curve pictured in Figure 6.41.
zS(x, y, z)ds.
Using these definitions, find the coordinates of the
center of mass of the wire.
7. Suppose that a wire is bent in the shape of a quarter
circle of radius a. Find the center of mass of the wire
if the density at points on the wire vary as the square
of the distance from the center of the wire.
8. (a) Find the centroid of the helical wire x = 3 cost,
y = 3 sinf, z = At, where 0 < t < 4jt. (Hint: No
calculation should be necessary.)
(b) Find the center of mass of the same wire if the den-
sity at each point of the wire is equal to the square
of the point's distance from the origin.
If a thin wire is bent in the shape of a curve C in the xy -plane
and has mass density at each point along the curve given by a
continuous function S(x, y), then we may define the moments
of inertia of C about the x- and y-axes, respectively, by
f
y S(x, y) ds,
Iy
L
x S(x, y) ds,
and the corresponding radii of gyration of C as
'(2,1,0)
Figure 6.41 The broken-line curve of
Exercise 4.
5. Find the average value of f(x, y, z) = z2 + xey on the
curve C obtained by intersecting the (elliptic) cylinder
x2/5 + y2 = 1 by the plane z = 2y.
6. A metal wire is bent in the shape of the semicircle
x2 + y2 = 4, y > 0, lying in the xy-plane. Suppose
that the mass density at each point (x, y, z) of the wire
is S(x, y, z) = 3 — y.
(a) Find the total mass of the wire.
(b) Using formulas analogous to those in §5.6, we de-
fine the (first) moments of the wire to be
/ xS(x,
Jc
y, z)ds,
j yS(x, y, z)ds,
ry
where M denotes the total mass M = fc S(x, y) ds of the wire.
Additionally, the moment of inertia of C about the origin (or,
equivalently, about the z-axis, if we think of the xy -plane as
embedded in R3,) is
j (x2 + y2)S(x,y)ds,
with corresponding radius of gyration r- = *JIZ/M. Exer-
cises 9-13 concern moments of inertia of curves.
9. (a) Consider the wire of Exercise 6 again. Find its mo-
ment of inertia about the y-axis.
(b) What is the radius of gyration r- for the wire about
the z-axis?
10. Find the moment of inertia Ix and the radius of gy-
ration rx about the x-axis of a straight wire between
(—2, 1) to (2, 3) whose density varies along the wire
as 8(x, y) = y.
1 1 . Find the moment of inertia lx and the radius of gyra-
tion rx about the x-axis of a wire shaped as the curve
y = x2 between (0, 0) and (2, 4) and whose density
varies as S(x, y) = x.
12. (a) Suppose that a thin metal wire is bent into a
curve C in R3 and has mass density at each point
452 Chapter 6 | Line Integrals
(x, y, z) along the curve given by a continuous
function S(x, y, z). Give general formulas analo-
gous to those in §5.6 for the moments of inertia of
the wire about the three coordinate axes.
(b) Find the moments of inertia about the coordinate
axes of a homogeneous (i.e., constant density)
wire shaped like the helix x = 3 cos t, y = 3 sin t ,
z = At, where 0 < t < Air. What are the radii of
gyration?
1 3. Find the moment of inertia /- about the z-axis of a wire
in the shape of the line segment between (— 1 , 1,2)
and (2, 2, 3) if the density along the segment varies as
S(x, y, z) = 1 + z2. What is rzl
14. Let r = f(6) be the polar equation of a curve in the
plane.
(a) Use scalar line integrals to show that the arclength
of the curve between (/(a), a) and (f(b), b) is
V(/(0))2 + (fWde.
r
(b) Sketch the curve r = sin (8 /2) and find its length.
15. (a) Give a formula in polar coordinates for the scalar
line integral of a function g(x, y ) along the curve
r = f(8), a <8 < b.
(b) Compute fc gds, where g(x, y) = x2 + y2 — 2x
and C is the segment of the spiral r = eie, 0 <
6 < 2ji.
Let C be a piecewise C , simple curve in R3. The total curva-
ture K of C is
K
j k ds,
where k denotes the curvature of C. (See §3.2 to review the no-
tion of curvature.) Exercises 16-20 involve the notion of total
curvature.
16. Show that if C is a simple curve of class C1
parametrized by x(r), a < t < b, then
J a
v x a
dt.
(Recall that v and a denote, respectively, the velocity
and acceleration of the path x.)
1 7. Find the total curvature of the helix
x(?) = (3 cost, 3 sinf, At), Q<t<\0n.
18. Find the total curvature of the parabola y = Ax2,
A > 0, for a < x < b.
19. Fenchel's theorem states that if C is a simple, closed
C1 curve in R3, then K > 2n and, moreover, K = In
if and only if C is a plane convex curve. (A simple,
closed curve C is convex if the line segment joining
any two points of C lies entirely in the region en-
closed by C.) Verify Fenchel's theorem for the ellipse
x2/a2 + y2/b2 = 1.
20. Let C be a simple, closed C1 curve in R3. Suppose that
the curvature k of C is bounded (i.e., 0 < k < I /a for
some a > 0).
(a) Show that if L denotes the length of C, then
L > 2na.
(b) What can you say about C if L = 2nal
21. Calculate the work done by the vector field F =
sin x i + cos yj+.xzkona particle moving along the
path x(r) = (t1, -t2, t), where 0 < t < 1.
22. Use Green's theorem to find the work done by the vec-
tor field F = x2y i + (x + y)y j in moving a particle
from the origin along the y-axis to the point (0, 1),
then along the line segment from (0, 1) to (1, 0), and
then from (1,0) back to the origin along the x-axis.
(Warning: Be careful.)
23. Use Green's theorem to recover the formula
f
for the area A of the region D described in polar coor-
dinates by
D = {(r, 6) | 0 < r < f(6), a <6 < b}.
24. Let C be a piecewise C , simple, closed curve in R2.
Show that
f{x)dx + g(y)dy = 0,
where / and g are any single-variable functions of
class C1.
25. Let D be a region in R2 whose boundary dD consists
of finitely many piecewise Cl, simple, closed curves
oriented so that D is on the left as you travel along any
segment of 3D. If (x, y) denotes the coordinates of the
centroid of D, show that
x = ^ £) x2 dy
2 ■ area of D J3D
and
1
area of D JdD
xy dy.
Also show that
and
1
area of D JdD
1
xy dx
y2 dx.
2 ■ area of D J3D'
26. Use the results of Exercise 25 to find the centroid of the
triangular region with vertices (0, 0), (1 , 0), and (0, 2).
Miscellaneous Exercises for Chapter 6 453
27. Use the results of Exercise 25 to find the centroid of
the region in R2 that lies inside the circle of radius 6
centered at the origin and outside the two circles of
radius 1 centered at (4, 0) and (—2, 2), respectively.
28.
Let C be a piecewise C1, simple, closed curve, ori-
ented counterclockwise, enclosing a region D in the
plane. Let n be the outward unit normal vector to D.
If f(x, y) and g(x, y) are functions of class C2 on D,
establish Green's first identity:
(/V2g + V/-Vg)dA
fVg-nds.
29. Under the hypotheses of Exercise 28, prove Green's
second identity:
(fV2g-gV2f)dA
(fVg-gVf)-nds.
A function g(x, y) is said to be harmonic at a point (xq, yo) if
g is of class C2 and satisfies Laplace 's equation
2 &h
dxL
+
dy2
0
on some neighborhood of(xo , yo). We say that g is harmonic on
a closed region OCR2 if it is harmonic at all interior points
ofD (i.e., not necessarily along dD). Exercises 30- 33 concern
some elementary results about harmonic functions in R2.
30. Suppose that D is compact (i.e., closed and bounded)
and that 3D is piecewise C1 and oriented as in Green's
theorem. Let n denote the outward unit normal vector
to 3D and let dg/dn denote Vg • n. (The term dg/dn
is called the normal derivative of g.) Use Green's first
identity (see Exercise 28) with f(x, y) = 1 to show
that, if g is harmonic on D, then
9g
dn
ds = 0.
31.
32.
Let / be harmonic on a region D that satisfies the
assumptions of Exercise 30. Show that
,.9/
V/-V/JA
dn
ds.
Suppose that f(x, y) = 0 for all (x, y) e 3D. Use Ex-
ercise 31 to show that f(x, y) = 0 throughout all of
D. (Hint: Consider the sign of V/ • V/.)
33. Let D be a region that satisfies the assumptions of Ex-
ercise 30. Use the result of Exercise 31 to show that if
fx and fi are harmonic on D and f\(x, y) = fi(x, y)
on 3D, then, in fact, f = fa on all of D. Thus, we see
that harmonic functions are determined by their values
on the boundary of a region. (Hint: Consider f\ — fi )
34. We call a vector field F on R3 radially symmetric if
it can be written in spherical coordinates in the form
F = f(p)ep, where ep is the unit vector that points in
the direction of increasing p-coordinate. (See §1.7.)
(a) Give an example of a (nontrivial) radially sym-
metric vector field, written in both Cartesian and
spherical coordinates.
(b) Show that if / is of class C1 for all p > 0, then
the radially symmetric vector field F = f(p)ep is
conservative.
-yi + xj
35. LetF
x2 + y2
(a) Verify Green's theorem over the annular region
D = {(x,y)|a2<Jc2-|-y2< 1}.
(See Figure 6.42.)
Figure 6.42 The annular region
of Exercise 35(a).
(b) Now let D be the unit disk. Does the formula of
Green's theorem hold for D? Can you explain why?
(c) Suppose C is any simple, closed curve lying out-
side the circle
Ca = {(x,y)\x2 + y2 = a2}
Figure 6.43 The curves C and C„ of
Exercise 35(c).
(See Figure 6.43 . ) Argue that if C is oriented coun-
terclockwise, then
x dy — y dx
x2 + y2
2jt.
454 Chapter 6 I Line Integrals
36. Consider the vector field F = — r i H — - j.
x2 + y2 x2 + y2
(a) Calculate V x F.
(b) Evaluate (fc F • ds, where C is the unit circle
x2 + y2 = 1.
(c) Is F conservative?
(d) How can you reconcile parts (a) and (b) with
Theorem 3.5?
37. (a) Let F = ey i + x4 j. Calculate the flux fc F • nds
of F across the boundary of the rectangle R =
[0, 1] x [0, 5].
(b) Let / and g be of class C1 and let F = f(y)i +
g(x ) j . Show that the flux of F across any piecewise
C1 simple, closed curve is zero.
38. Use Newton's second law of motion F = ma. to show
that the work done by a force field F in moving a par-
ticle of mass in along a path x(f ) from x(a) to x(b) is
equal to the change in kinetic energy of the particle. In
other words,
j^F-d$ = \m{v(b))2 - \m(v(a))2,
where v(t) = \\v(t)\\ , the speed at time t. (Use the prod-
uct rule for dot products of vector- valued functions.)
39. Let F be a conservative vector field on R3 with F =
— VV. If a particle travels along a path x, recall that
its potential energy at time t is defined to be V(x(f)).
Use line integrals to prove the law of conservation of
energy: As a particle of mass m moves between any
two points A and B in a conservative force field, the
sum of the potential and kinetic energies of the particle
remains constant. (Use Exercise 38 and Theorem 3.3.)
The use of line integrals provides an alternative proof
of Theorem 4.2 in §4.4.
Surface Integrals
and Vector Analysis
7.1 Parametrized Surfaces
7.2 Surface Integrals
7.3 Stokes's and Gauss's
Theorems
7.4 Further Vector Analysis;
Maxwell's Equations
True/False Exercises for
Chapter 7
Miscellaneous Exercises for
Chapter 7
7.1 Parametrized Surfaces
Introduction
Surfaces in R3 may be presented analytically in different ways. Here are two
familiar descriptions:
1. as a graph of a function of two variables, that is, as points (x, y, z) in R3
satisfying z = f(x, y) (e.g., z = x2 + Ay2);
2. as a level set of a function of three variables, that is, as points (x, y, z)
such that F(x, y,z) = c for some suitable function F and constant c (e.g.,
z2y5
7
x y
-2-5 + x = 1).
Both of these descriptions are problematical. As noted in §2.1, many common
surfaces cannot be described as graphs of functions of two variables. Recall, for
example, that the full sphere x2 + y2 + z2 = 1 is not the graph of a function of
two variables. Therefore, description 1 is not sufficiently general.
There are also problems with description 2. Not all equations of the form
F(x, y,z) = c have solutions that fill out surfaces. Indeed although the level set
of F(x, y, z) = x2 + y2 + z2 at height 1 is a sphere, at height 0 it is a single point,
and at height — 1 completely empty. In addition, it is somewhat tricky to describe
surfaces (i.e., two-dimensional objects) in R" by using level sets when n is larger
than 3. Another approach is desirable for presenting surfaces analytically, in order
to avoid the problems just mentioned, to emphasize clearly the two-dimensional
nature of a surface and to facilitate subsequent calculations. With this discussion
in mind, we state the following definition:
DEFINITION 1 .1 Let D be a region in R2 that consists of a connected open
set, possibly together with some or all of its boundary points. A parametrized
surface in R3 is a continuous function X:flcR2-^R3 that is one-one on
D, except possibly along 3D. We refer to the image X(D) as the underlying
surface of X (or the surface parametrized by X) and denote it by S. (See
Figure 7.1.)
The restrictions on the region D and the map X of Definition 1.1 are meant to
ensure that D is a two-dimensional subset of R2 with a two-dimensional image.
456 Chapter 7 | Surface Integrals and Vector Analysis
c
X(D) = S
Figure 7.1 A parametrized surface.
If we write the component functions of X, then, for (s, ?) e D,
X(s, t) = (x(s, ?), y(s, ?), z(s, ?)),
and the underlying surface S can be described by the parametric equations
x = x(s, ?)
y = y(s, ?) (s, t) e D.
z = z(s, t)
(1)
EXAMPLE 1 Consider the parametrized surface X: R2
X(*,0 = .(i-j) + f(i + 2k) + 3j.
R described by
The image of X, shown in Figure 7.2, is the plane through the point (0, 3, 0),
determined by the vectors a = i — j and b = i + 2k. (See Proposition 5.1
of §1.5.) ♦
(0,3,0) Y
= X(0, 01. v
Figure 7.2 The parametrized plane of Example 1.
EXAMPLE 2 Let D = [0, 2tt) x [0, n] and consider X: D -> R3 given by
X(s, t) = (a coss sin?, a sins sin/, a cos?).
The corresponding parametric equations are
x = a cos s sin /
y = a sins sin? 0 < s < 2n, 0 < ? < n.
z = a cos ?
The parametric equations imply that x2 + y2 + z2 = a2, meaning all the points
of S = X(D) lie on a sphere of radius a centered at the origin. The paramet-
ric equations are precisely the spherical-rectangular coordinate conversions (see
§1.7) with the p -coordinate held constant at a and with s and ? used instead of 6
and (p. Hence, the image of X is indeed all of the sphere. (See Figure 7.3.) ♦
7.1 j Parametrized Surfaces 457
t z
Image of
Figure 7.3 A sphere rendered as a parametrized surface.
EXAMPLE 3 The points of the surface parametrized by
x = a cos s
■ y = a sin s 0 < s < 2jt
z = t
satisfy the equation x2 + y2 = a2 and so can be seen to form an infinite cylinder
of radius a. Figure 7.4 shows how the function X(s, /) = (a coss, a sins, t) maps
the infinite strip D = {(s, t) | 0 < s < 2n] onto the cylinder by "gluing" the line
s = 0 to s = 2ir . ♦
Figure 7.4 The map X glues together the edges of D to form
a cylinder.
EXAMPLE 4 Let D c R2 be an open set, possibly together with some or all of
its boundary points. If /: D -> R is a continuous scalar- valued function of two
variables, then it is not difficult to parametrize the graph of /: We let
R3
X: D -> R3 be
That is, the parametric equations
X(s, t) = (s, t, f(s, 0).
0, 0 e D
describe the points of the graph of /. (See Figure 7.5.)
x = s
y = t
z = f(s, t)
Coordinate Curves, Normal Vectors,
and Tangent Planes
Let S be a surface parametrized by X: D —> R3. If we fix / = to and let only s
vary, we obtain a continuous map
s i — > X(j, to),
458 Chapter 7 i Surface Integrals and Vector Analysis
D
X(£>) is graph of z =f(x, y)
Figure 7.5 The graph of z = f(x, y) as a parametrized surface.
whose image is a curve lying in S. We call this curve the s-coordinate curve at
t = t0. Similarly, we may fixj = i0 and obtain a map
t i — > X(s0, t),
whose image is the f -coordinate curve at s = so- Figure 7.6 suggests the appear-
ance of the coordinate curves.
s = s0
7 f
D
t = tn
X(M„)
X(s„,0
Figure 7.6 The coordinate curves of a parametrized surface.
EXAMPLE 5 The points of the parametrized surface T defined by
X = (a + & COS 0 COS J n^.^o
; , , : . 0 < s < 2tt, 0 < t < 2jt:
y = (a + ocosr)sms ~ _ . ~ ,
, . a, b positive constants with a > b,
z = b sin t F
satisfy the equation
(y/x2 + y2-ay + z2=b2.
The ^-coordinate curve at t = 0 is
x = (a + b) cos s
V = (a + b) sin s
z = 0
and is readily seen to be the circle of radius a + b, centered at the origin and lying
in the xy-plane. In general, you may check that the s-coordinate curve at t = t0
is a circle of radius a + b cos to (which varies between a — b and a + b) in the
horizontal plane z = b sinfo- (See Figure 7.7.)
7.1 | Parametrized Surfaces 459
s-coordinate
curve at t = kI2
s-coordinate
curve at t = 0
s-coordinate
curve at t = n
Figure 7.7 Some s-coordinate curves for the torus T of Example 5.
The f-coordinate curve at s = 0 is
x = a + b cos f
y = 0
z = bsmt
The image is a circle of radius b centered at (a, 0, 0) in the xz-plane (i.e., the
plane y = 0). At s = so, the f-coordinate curve is
x = cos so {a + b cos t)
y = sin so (a + b cos t) .
z = bsint
Along this curve, we have y/x = tans0, a constant. The curve lies in the vertical
plane (sin so)x — (cos so)y = 0. Moreover, it is not hard to see that the distance
from any point on this curve to the point P(a cos so, a sin so, 0) is b, and the
image is a circle of radius b centered at P. See Figure 7.8 for examples of
f-coordinate curves.
s = 3nl2
s = nl2
kI2 k 3nl2 2n
Figure 7.8 Some f-coordinate curves for the torus T of Example 5.
The aforementioned surface T is called a torus, a doughnut-shaped surface
shown in Figure 7.9. It is generated both by the collection of the s-coordinate
curves and by the collection of the f-coordinate curves. ♦
Suppose that X(s, f) = (x(s, f), y(s, f), z(s, f)), where (s, f) 6 D, is a dif-
ferentiable (or C1) map, in which case we say that the parametrized surface
S = X(D) is differentiable (or C1) as well. Then the coordinate curves X(s, fo)
and X(so, f) have well-defined tangent vectors at points (so, fo) in D. (See
Figure 7.10.) To find the tangent vector Ts to the s-coordinate curve X(s, fo)
460 Chapter 7 | Surface Integrals and Vector Analysis
5-coordinate
/ curve at t = t0
Figure 7.9 The parametrized torus T.
X(s0,t)
Figure 7.10 The tangent vectors Ts and Tf to the coordinate curves.
at (so, to), we differentiate the component functions of X with respect to s and
evaluate at (sq, to):
Ts(s0, t0) = —(s0, to) = t-(«o, *o) i + t-(«o, to)] + —(so, t0) k. (2)
as as as as
Similarly, the tangent vector T, to the r-coordinate curve X(so, t) at (s0, t0) may
be calculated by differentiating with respect to /:
3X dx dy dz
Tt(s0, to) = —(so, to) = —(so, t0)i+ —(so, to)} + —(so, fo)k. (3)
at at at at
Since Ts and T, are both tangent to the surface S at (so, to), the cross product
Ts(so, to) x Tt(so, to) will be normal to S at (so, to), provided it is nonzero.
DEFINITION 1.2 The parametrized surface S = X(D) is smooth at
X(so, to) if the map X is of class C1 in a neighborhood of (so, to) and if
the vector
N(50, = T,(j0, h) x T,(i0, to) + 0.
If 5 is smooth at every point X(so, to) € S, then we simply refer to S as a
smooth parametrized surface. If S is a smooth parametrized surface, we call
the (nonzero) vector N = T, x T, the standard normal vector arising from
the parametrization X.
7.1 ; Parametrized Surfaces 461
EXAMPLE 6 We claim that the torus T of Example 5 is smooth. Recall that T
is given as the image of the map
X: [0,2tv] x [0,2*] -> R3,
X(s, /) — ((a + b cos / ) cos s, (a + b cos /) sin s , b sin /),
where a > b > 0. Then from formulas (2) and (3), we have
TsC?o> fo) = —(fl + &cos/o) sins0 i + (a + cos/o)cosso j
and
T((so> *o) = —b sin/0 coss0 i — b sin/0 sini0 j + bcosfo k,
so that
Ts x T, = (a + b cos fo)^ cos to cos so i + (a + ^ cos ?o)^ cos /o sin so j
+ (a + b cos fo)^ sin t0 k
= + b cosfo)(cos?o coss0 i + cosfo sinso j + sinf0 k).
Since a > b > 0, the factor b(a + b cos fo) is never zero. Furthermore, since the
sine and cosine functions are never simultaneously zero, at least one component
of Ts x T, is never zero. Hence, the torus is a smooth parametrized surface. ♦
s = —2 t s = 2 z
t = n/4
Figure 7.1 1 The cone z2 = x2 + y2 as a parametrized surface.
EXAMPLE 7 The equation z2 = x2 + y2 defines a cone in R3. (See Figure
7.11.) If z is held constant (which corresponds to slicing the surface by a
horizontal plane), then the expression x2 + y2 is constant. Hence, the constant-z
cross sections are circles of radius |z| or a single point in the case of the vertex.
This suggests that we can parametrize the cone by using one parameter variable
for z and another for the angle around the z-axis. Thus, we obtain the following
equations:
X = s cos t
■ y = s sin/ 0<f< 2tt.
z = s
Then we have
T, . = cos/ i + sinf j + k and T, = — s sin/ i + s cos t j.
462 Chapter 7 | Surface Integrals and Vector Analysis
Therefore,
TsxT,=
i J
cos t sin t
-ssint scosf
k
1
0
= — s cos t i — s sin t j + s k.
Note that Ts x T, = 0 when (and only when) s = 0. The cone fails to be smooth
just at its vertex (the single point of the underlying surface corresponding to
s = 0). ♦
Examples 6 and 7 suggest why the terminology "smooth" is used: Intuitively,
a parametrized surface is smooth at a point if it has no sharp "cusps" or "corners"
there. This is true of the torus but not of the cone, which has a singularity at its
vertex.
If a parametrized surface is smooth at a point X(so, to), then we define the
tangent plane to S = X(D) at the point X(so, to) to be the plane that passes
through X(s0, to) and has
N($0, to) = Ts(so, to) x Tt(s0, to)
as normal vector. To write an equation for this plane, we denote (x, y, z) by the
(variable) vector x. Then the tangent plane equation is
N(s0,fo)-(x-X(so,fo)) = 0.
(4)
If we write the components of N(so, to) as Ai + B j + Ck and X(s0, to) as
(x(s0l t0), y(s0, to), z(so, to)), then we may expand equation (4) to obtain
A(x - x(s0, to)) + B(y - y(s0, t0)) + C(z - z(s0, t0)) = 0. (5)
EXAMPLE 8 Consider once again the parametrized cone of Example 7 as
X(s, t) = (s cost, s sinf, s).
From the calculations in Example 7, the cone is smooth at the point (0, 1,1) =
X(l, n/2), and so the tangent plane exists at that point. We have
(l,f)=j + k and T,(l,f)
so that
N
= T,(l,f)xTf(l,!) =
i
0
-1
= -j + k.
Hence, equation (5) can be applied to verify that an equation for the tangent
plane is
or, more simply,
0(x-0)-l(y-l) + l(z-l) = 0
z = y.
EXAMPLE 9 If f(x, y) is of class C1 in a neighborhood of a point (x0, y0)
in its domain D, then the graph of / is a smooth parametrized surface at
(xo, yo, f(xo, yo))- Recall from Example 4 that the graph of / is parametrized by
7.1 Parametrized Surfaces 463
X: D -> R3,
X(s, t) = (s, t, f(s, 0). Then
Ts = i H k and T, = j +
3s
so that
N = Ts x T, = - — i - — j + k.
ds dt3
Note that N is nonzero at any point (s, t, f(s, t)) = (x, y, fix, y)).
Next, consider the surface defined as the level set
S = {(x, y, z) | F(x, y, z) = c, c constant}.
z
Figure 7.1 2 A cube.
If F is of class C1 in a neighborhood of a point (x0, Vo, Zo) £ S and VF(x0, yo,
z0) ^ 0, then the implicit function theorem (Theorem 6.5 of §2.6) implies that, in
principle at least, the defining equation F(x, y, z) = c of S always can be solved
locally for (at least) one of the variables x, y, or z in terms of the other two.
In other words, under the given assumptions on F, the level set S is locally the
graph of a C1 function of two variables. It is important to remember that the
idea of "solving locally" does not mean that all points of S can be described as
the graph of a single function (as we already know quite well in the case of the
sphere*2 + y2 + z2 = 1, for example), but rather that, near points (xo, yo, Zo) £ S
where V F(xq, yo, zo) ^ 0, a portion of S may be described as a graph. Hence,
graphs of C1 functions of two variables and level sets of C1 functions of three
variables, under certain smoothness hypotheses, are locally equivalent descrip-
tions for surfaces. Moreover, since the graph of a C1 function is a smooth
parametrized surface, we may shift relatively freely among our three descriptions
for surfaces. ♦
Smooth parametrized surfaces are of primary importance because of the ease
with which we may adapt techniques of calculus (particularly integral calculus)
to them. But we are also interested in piecewise smooth parametrized surfaces.
DEFINITION 1 .3 A piecewise smooth parametrized surface is the union of
images of finitely many parametrized surfaces X, : D;- — > R3, i = 1, . . . , m,
where
• Each Di is a region in R2 consisting of a connected open set, possibly
together with some of its boundary points (for the most part, we want Dt
to be an elementary region);
• Each X, is of class C1 and one-one on all of D, , except possibly along
9 A;
• Each Sj = X, (D, ) is smooth, except possibly at finitely many points.
EXAMPLE 10 The surface of a cube is a piecewise smooth parametrized
surface. It is the union of its six faces, each one of which is a smooth parametrized
surface, namely, a portion of a plane.
More explicitly, suppose that a cube's faces are portions of the planes x = 0,
x = 1, y = 0,y = l,z = 0,andz = 1 as in Figure 7. 12. Then we may parametrize
464 Chapter 7 | Surface Integrals and Vector Analysis
the cube's faces by X, : [0, 1] x [0, 1] -> R3, = 1, . . . , 6, where
Xl(s,t) = (0,s,t); X2(s,t) = (\,s,t); X3(s,t) = (s,0,t);
X4(s,t) = (s,l,t); X5(s,t) = (s,t,0); X6(s,t) = (s,t,l).
Each map X, is clearly of class C1 and one-one. In addition, the faces have
well-defined nonzero normal vectors. For example, for both Xi and X2,
N1=N2 = T5xT(=jxk = i.
Similarly,
N3 = N4 = i x k = -j and N5 = N6 = i x j = k.
None of these vectors vanishes. There is no consistent way to define normal
vectors along the edges of the cube (where two faces meet). That is why the cube
is only piecewise smooth. ♦
Area of a Smooth Parametrized Surface
Now, we use the notion of a parametrized surface to calculate the surface area of a
smooth surface. In the discussion that follows, we take S = X(D) to be a smooth
parametrized surface, where D is the union of finitely many elementary regions
in R2 and X: D -> R3 is of class C1 and one-one except possibly along 3D.
x
Figure 7.1 3 The image of the As x At rectangle in D is approximately a
parallelogram spanned by Ts(sq, to) As and T,(sq, to) At.
The key geometric observation is as follows: Consider a small rectangular
subset of D whose lower left corner is at the point (s0, t0) e D and whose width
and height are As and Af , respectively. The image of this rectangle under X is a
piece of the underlying surface S that is approximately the parallelogram with a
corner at X(so, to) and spanned by the vectors Ts(so, to) As and T,(so, to) At. (See
Figure 7.13.) The area A A of this piece is
AA % ||Ts(s0, f0)As x Tf(i0, t0)At\\ = \\Ts(s0, f0) x T,(s0, t0)\\AsAt.
Now, suppose D = [a, b] x [c, d]; that is, suppose D itself is a rectangle.
Partition D into n2 subrectangles via
a = so < s\ < ■ ■ ■ < sn = b and c = to < t\ < ■ ■ ■ < t„ = d.
Let Asi = Sj — and Af, = tj — tj-\ for i, j — 1, . . . , n. Then S is in turn
partitioned into pieces, each of which is approximately a parallelogram, assuming
As,- and Atj are small for i, j = 1, . . . , n. If AAy denotes the area of the piece
7.1 Parametrized Surfaces 465
of S that is the image of the ijth subrectangle of D, then
n
Surface area of S = ^ AA,;
n
« J] ||T,(*,_i,*;_i)xT,(j/_1,fy_i)||Aj/A^.
Therefore, it makes sense to define
Surface area of S = lim > AAy = / / ||TsxT, || rfsrfr
and, in general, where D is an arbitrary region (i.e., not necessarily a rectangle),
Surface area of S = I I ||TS x T, || Jr. (6)
Formula (6) can be extended readily to the case where 5 is a piecewise smooth
parametrized surface by breaking up the integral in an appropriate manner.
EXAMPLE 1 1 We use formula (6) to calculate the surface area of a sphere of
radius a. Recall from Example 2 that the map
X(s, t) = (a coss sinf, a sins sinf, a cos t), 0 < s < 2it, 0 < t < it
parametrizes the sphere in a one-one fashion. Then
Ts = —a sin s sin t i + a cos s sin t j
and
T, = a cos s cos t i + a sin s cos f j — a sin f k.
Hence,
so that
Tj xT, = — a2 sin f (cos s sin f i + sin s sin f j + cos / k),
\TS x T, || = a2 sin?.
Therefore, formula (6) becomes
r f2n ? r ?
Surface area = J J a sin/ ds dt — I 2na sintdt
Jo Jo Jo
= 2ttci2(— cos t)\" = 2na2(l + 1) = Ana2.
This result checks with the well-known formula for the surface area of a sphere.
Note, however, that if we let 0 < s < 4jt, 0 < t < ir, then the image of X is the
same sphere, but formula (6) would yield
-Ait
a2 sin? ds dt = in a2.
P1Z
Jo Jo
This "overcount" is why we must assume that the map X is (nearly) one-one.
(Note: The aforementioned parametrization fails to be smooth at t = 0 and at
t = jt, but this is along dD and so does not affect the surface area integral.) ♦
466 Chapter 7 i Surface Integrals and Vector Analysis
If we write the component functions of X as
X(s, f) = (x(s, t), y(s, t), z(s, 0),
we find that
Ts x T, =
i
j
k
dx
Oy
dz
97
3 s
3s
dx
dy
dz
It
dt
97
dy dz dy dz\ . ( dx dz dx dz
ds ~dl ~ ~dtds) 1+ \ dlYs ~ JsYt
Using the notation of the Jacobian, we obtain
d(x,z)
J +
dx dy
97 97
dx dy
97 97
NO, t) = Ts x T,
, d(x,y)
J + — rk.
(7)
90,0 90,0" ' 30,0
This alternative formula for the normal vector to a smooth parametrized surface
will prove useful to us on occasion. For the moment, we take its magnitude:
UNO, Oil =
+
9(x, z)
90, 0 / V 30, 0
Hence, formula (6) may also be written as
+
9(y, z)
30, 0
EXAM PLE 1 2 Find the surface area of the torus described in Example 5 . Recall
that the torus is parametrized as
x = (a + b cos 0 cos s
y = (a + bcost)sins 0<s,t<2n, a > b > 0.
z = b sin t
Thus,
3Q, y)
90, 0
— (a + Z? cos 0 sins
(a +b cos f ) cos ^
-bsint coss
sin? sin s
(a + b cos f )(£> sin / sin2 s + bsint cos2 s)
d(x, z)
30, 0
(a + Z? cos O^7 sint,
-(a + Z? cos 0 sins
0
-bsint cos 5
Z?cosr
= —(a + b cos 0& cos t sins,
7.1 | Exercises 467
and
3(y, z)
(a + &cosf)coss
0
3(5, f)
= (fl + Z? cos cos f COS 5
By formula (8), we have
Surface area
bsint sin s
b cos r
JO JO
(a + b cos r)2[Z?2 sin2 f + b2 cos2 / sin2 s + b2 cos2 f cos2 .?] ds dt.
Using the trigonometric identity cos2 8 + sin2 6 = 1 twice, we simplify the inte-
gral to
p 2jr p 2.71 p 2tc
I I (a + b cos t)b ds dt = I 2nb(a + b cos t)dt
Jo Jo Jo
= 2jib{at + b sinf)^"
= 4n2ab. ^
EXAMPLE 1 3 Suppose that a smooth surface is described as the graph of a C1
function /(x, y), that is, by the equation z = f(x, y), where (x, y) varies through a
plane region/). Then the standard parametrization X(.5\ /) = (s, t, f(s, r))implies
T.s x T, = -fsi- ft j + k.
(See Example 9.) Formula (6) yields
Surface area = j j \\TS x T(|| ds dt = j j J f2 + / 2 + 1 ds dt.
Since x = s, y = t in this parametrization of the graph, we conclude that
Surface area of the graph of f(x, y) over D
-IL
f2 + f2 + ldxdy.
(9)
One final note: It is not at all clear that either formula (6) or formula (8)
depends only on the underlying surface S = X(D) and not on the particular
parametrization X. These formulas are independent of the parametrization, as we
shall observe in the following section, in the context of general surface integrals.
7.1 Exercises
1 . Let X: R2 —> R3 be the parametrized surface given by (b) Find an equation for the plane tangent to this surface
, 9 ? 9 „ ^ at the point (3, 1, 1).
X(s,t) = (s2 -t2,s + t,s2 + 3f). F v '
2. Find an equation for the plane tangent to the torus
(a) Determine a normal vector to this surface at the
point = ^ ^ cos t) cos s, (5 + 2 cos t) sin s, 2sinf)
(3, 1,1) = X(2, -1). at the point ((5 - V3)/V2. (5 - V3)/V2, l).
468 Chapter 7 | Surface Integrals and Vector Analysis
3. Find an equation for the plane tangent to the surface
X = es, y = t2e2s, z = 2e~s + 1
at the point (1,4,0).
4. Let X(s, t) = (s2 cost, s2 sinf, s), -3 < s < 3, 0 <
t < In.
(a) Find a normal vector at (s, t ) = (— 1 , 0).
(b) Determine the tangent plane at the point (1, 0, —1).
(c) Find an equation for the image of X in the form
F(x,y,z) = Q.
5. Consider the parametrized surface X(s, f) = (s, s2 +
t,t2).
(a) Graph this surface for — 2 < s < 2, —2 < t < 2.
(Using a computer may help.)
(b) Is the surface smooth?
(c) Find an equation for the tangent plane at the point
(1,0,1).
6. Describe the parametrized surface of Exercise 1 by an
equation of the form z = f(x,y).
7. Let S be the surface parametrized by x = scost,
y = s sinf, z = s2, where s > 0, 0 < f < 2n .
(a) At what points is S smooth? Find an equation for
the tangent plane at the point ( 1 , V3, 4).
(b) Sketch the graph of S. Can you recognize S as a
familiar surface?
(c) Describe S by an equation of the form z = f(x, y).
(d) Using your answer in part (c), discuss whether S
has a tangent plane at every point.
8. Verify that the image of the parametrized surface
X(s, t) = (2 sin s cos f , 3 sins sinf, coss),
0 < s < TV, 0 < f < 2jt,
is an ellipsoid.
9. Verify that, for the torus of Example 5 , the s -coordinate
curve, when f = to, is a circle of radius a + b cos to.
1 0. The surface in R3 parametrized by
X(r, 6) = (rcos6,rsia6, 9), r>0, -oo<6<oo,
is called a helicoid.
(a) Describe the r-coordinate curve when 0 = jr/3.
Give a general description of the r -coordinate
curves.
(b) Describe the ^-coordinate curve when r = 1.
Give a general description of the 8 -coordinate
curves.
(c) Sketch the graph of the helicoid (perhaps using
a computer) for 0 < r < 1 , 0 < 9 < An . Can you
see why the surface is called a helicoid?
11. Given the sphere of radius 2 centered at (2, —1, 0),
find an equation for the plane tangent to it at the point
(1, 0, V2) in three ways:
(a) by considering the sphere as the graph of the
function
fix, y) = j4-{x-2)2-{y+\)2;
(b) by considering the sphere as a level surface of the
function
F(x, y, z) = (x- 2)2 + (y + l)2 + z2;
(c) by considering the sphere as the surface para-
metrized by
X(s, t ) = (2 sin s cos f + 2, 2 sin s sin f — 1,2 cos s ).
In Exercises 12—15, represent the given surface as a piecewise
smooth parametrized surface.
12. The lower hemisphere x2 + y2 + z2 = 9, including the
equatorial circle.
13. The part of the cylinder x2 + z2 = 4 lying between
y = — 1 and y = 3.
14. The closed triangular region in R3 with vertices
(2,0, 0), (0, 1,0), and (0,0, 5).
15. The hyperboloid z2 — x2 — y2 = 1. (Hint: Use two
maps to parametrize the surface.)
16. This problem concerns the parametrized surface
X(s,t) = (s3,t3,st).
(a) Find an equation of the plane tangent to this surface
at the point (1, -1, -1).
(b) Is this surface smooth? Why or why not?
(c) Use a computer to graph this surface for — 1 < s <
1, -1 < t < 1.
(d) Verify that this surface may also be described by
the xyz-coordinate equation z = ffxy. Try using
a computer to graph the surface when described in
this form. Many software systems will have trou-
ble, or will provide an incomplete graph, which is
one reason why parametric descriptions of surfaces
are desirable.
17. The surface given parametrically by X(s, t ) =
(st,t,s2) is known as the Whitney umbrella.
(a) Verify that this surface may also be described by
the jcyz-coordinate equation y2z = x2.
(b) Is X smooth?
(c) Use a computer to graph this surface for —2 < s <
2, -2 < f < 2.
(d) Some points (x, y, z) of the surface do not cor-
respond to a single parameter point (s, f). Which
ones? Explain how this relates to the graph.
(e) Give an equation of the plane tangent to this sur-
face at the point (2, 1, 4).
7.2 | Surface Integrals 469
(f) Show that at the point (0, 0, 1) on the image of X
it's reasonable to conclude that there are two tan-
gent planes. Give equations for them.
18. Let S be the surface defined as the graph of a function
f(x, y) of class C1. Then Example 4 shows that S is
also a parametrized surface. Show that formula (5) for
the tangent plane to S at (a , b , f(a , b)) agrees with that
of formula (4) in §2.3.
1 9. (a) Write a formula for the tangent plane to a surface
described by the equation y = g(x, z).
(b) Repeat part (a) for a surface described by the equa-
tion x = h(y, z).
20. Suppose X: D — > R3 is a parametrized surface that is
smooth at X(sq, to). Show how the definition of the
derivative DX(sq, to) (see Definition 3.8 of Chapter 2)
can be used to give vector parametric equations for the
plane tangent to S = X(D) at the point X(sq, to).
21 . Use the result of Exercise 20 to provide parametric
equations for the plane tangent to the surface X(s, t ) =
(s, s2 + t, t2) at the point (1,0, 1). Verify that your an-
swer is consistent with that of Exercise 5(c).
22. Use the parametrization in Example 3 to verify that the
surface area of a cylinder of radius a and height h is
2nah.
23. Let D denote the unit disk in the jf-plane. Let X: D — >
R3 be defined by (s + t, s — t , s ). Find the surface area
ofX(D).
24. Find the surface area of the helicoid
X:D^R3, X(r, 6) = (r cos0,r sin 6, 6)
for 0 < r < 1, 0 < 9 < 2nn, where n is a positive
integer.
25. A cylindrical hole of radius b is bored through a ball
of radius a (> b ) to form a ring. Find the outer surface
area of the ring.
26. Find the area of the portion of the paraboloid z =
9 — x2 — y2 that lies over the xy-plane.
27. Find the area of the surface cut from the paraboloid
z = 2x2 + 2y2 by the planes z = 2 and z = 8.
28. Calculate the surface area of the portion of the plane
x + y + z = a cut out by the cylinder x2 + y2 = a2 in
two ways:
(a) by using formula (6);
(b) by using formula (9).
29. Let S be the surface defined by the equation z =
f(x, y). If f2 + f2 = a, where a is a positive con-
stant, determine the surface area of the portion of S
that lies over a region D in the .ty-plane in terms of the
area of D.
30. Let S be the surface defined by
1
z = , for z > 1 .
(a) Sketch the graph of this surface.
(b) Show that the volume of the region bounded by S
and the plane z = 1 is finite. (You will need to use
an improper integral.)
(c) Show that the surface area of S is infinite.
31 . Find the surface area of the intersection of the cylinders
x2 + y2 = a2 and y2 + z2 = a2.
32. Suppose that a surface is given in cylindrical coordi-
nates by the equation z = f(r, 0), where (r, 6) varies
through a region D in the r#-plane where r is non-
negative. Show that the surface area of the surface is
given by
33. Suppose that a surface is given in spherical coordi-
nates by the equation p = f(ip, 6), where (</s, 9) varies
through a region D in the ^6*-plane and f(<p, 9) is non-
negative. Show that the surface area of the surface is
given by
x 7(/(?, 9)2 + Mcp, 9)2) sin2 <p + f0(<p, 9)2 d<p d9.
7.2 Surface Integrals
In this section, we will learn how to integrate both scalar-valued functions and
vector fields along surfaces in R3. We proceed in a manner that is largely analo-
gous to our explorations of line integrals in §6.1: We begin by defining suitable
integrals over parametrized surfaces and then establish that the particular choice of
parametrization doesn't much matter — that, really, only the underlying surface is
important, and possibly the orientation.
470 Chapter 7 I Surface Integrals and Vector Analysis
Area AAt
Figure 7.1 4 A small piece
of the surface S has area
A At. The point is located
in this surface piece.
Scalar Surface Integrals
Suppose 5 is a bounded surface in R3 and f(x , y, z) is a continuous, scalar- valued
function whose domain includes S. Then we want the surface integral ffs f dS to
be a limit of some kind of Riemann sum. So suppose S is partitioned into finitely
many small pieces and that the area of the kth piece is AA*. Let c* denote an
arbitrary "test point" in the kth piece. (See Figure 7. 14.) Then the surface integral
of / over S should be
fffds= „1t nE/(c*)AA*
(1)
provided of course, that this limit exists.
Now, we add some formalism to provide a proper definition. Suppose that
S is a smooth parametrized surface; that is, suppose that S is the image of the
C1 map X: D — >• R3, where D is a connected, bounded region in R2. Let / be a
continuous function defined on S = X(D). As seen in §7.1, the small rectangle
in D, having dimensions As and At with lower left corner at the point (sq, to), is
mapped by X to a piece of surface that is approximately a parallelogram of area
AA sa ||T,Oo, to) x T,0o, t0)\\AsAt.
(See Figure 7.13, page 464.) Suppose D is the rectangle [a, b] x [c, d] and that
we partition D by
a = sq < S\ < ■ ■ ■ < s„ = b and c = to < t\ < ■ ■ ■ < tn = d.
Then the limiting sum that formula (1) represents is
lim T f(X(s*, t*)) ||T,0,_i, x T/(s/_i, Asj Atj, (2)
all Ajj.Afi-^O r—1,
i,j=l
where
Asi = Si - Si-i, Atj = tj - tj-i, Si-i < s* < Si, and t}-\ < t* < tj.
(Thus, X(**, tj) is an arbitrary point in the image of the subrectangle [si-i, st] x
[tj-i, tj] and, hence, a "test point" in the corresponding small surface piece.) But
then the limit in formula (2) is
f f f(X(s,t))\\JsxJt\\dsdt.
J c J a
When D is a more arbitrary region than a rectangle, it makes sense to use the
following definition for the surface integral of a function over a parametrized
surface:
DEFINITION 2.1 Let X: D — > R3 be a smooth parametrized surface, where
D c R2 is a bounded region. Let / be a continuous function whose domain
includes S = X(D). Then the scalar surface integral of / along X, denoted
JfxfdS, is
j 'f fdS = ff f(X(s, t)) \\TS x T, || ds dt
-IL
f(X(s,t)) \\N(s,t)\\dsdt.
7.2 | Surface Integrals 471
Although we need not assume that the map X is one-one on D in order to
work with the integral in Definition 2.1, in practice we usually find it useful to
take X to be one-one, except perhaps along dD. If this is the case, and if / is
identically 1 on all of X(£)), then
if fdS= if ldS= if || Ts x Tf \\dsdt = surface area of X(D),
J Jx J Jx J Jd
as stated by formula (6) in §7. 1 . The scalar surface integral in Definition 2. 1 is thus
a generalization of the integral we use to calculate surface area. We can think of
ffxfdS as the limit of a "weighted sum" of surface area pieces, the weightings
given by /. If / represents mass or electrical charge density, then ffxfdS yields
the total mass or total charge on X(D) (assuming X is one-one, except perhaps
along dD).
For computational purposes, recall that if we write the components of X as
X(s, t) = (x(s, t), y(s, t), z(s, t)),
then
™ t\ t ,t d(y>z). d(x,z). d(x,y)
N(i, t) = To x T, = i ■ H k.
v ' d(s, t) d(s, t) J d(s, t)
We obtain
f(x(s, t), y(s, t), z(s, 0)
d(s, t) J V 90> 0 / V 9(s, t)
EXAMPLE 1 We evaluate / fx z3 dS, where X: [0, 2tt] x [0, n] -+ R3 is the
parametrized sphere of radius a :
X(s, /) = (a coss sinf, a sins sinf, a cos t).
Using Definition 2.1 or its reformulation in formula (3), we find that
m°-m = i\jM) +{hJT)) +l«Z7)j
= J a4(sin2 t cos2 t + sin2 s sin4 1 + cos2 s sin4 1)
= a2\J sin2 / cos2 t + sin4 1
= a2J sin2 ? (cos2 t + sin2 f)
= a2 sinf.
(See also Example 1 1 in §7.1.) Hence,
'jr pin
/> /> ptx p Lit p jr
/ / z3 dS = I I (a cost)3a2 sint ds dt = a5 / 2tt cos3 t sin f
= 27rfl5 Kcos4f)|o =2na5 (_I-(-I))=0.
472 Chapter 7 | Surface Integrals and Vector Analysis
S*. z = 15
Sl:x2 + y1 = 9
/ S2:z = 0
x
Figure 7.15 The closed
cylinder of radius 3 and
height 15 of Example 2.
To define and evaluate scalar surface integrals over piecewise smooth para-
metrized surfaces, simply calculate the surface integral over each smooth piece
and add the results.
EXAMPLE 2 Let 5 be the closed cylinder of radius 3 with axis along the z-axis,
top face at z = 15, and bottom face at z — 0, as shown in Figure 7.15. Then 5
is a piecewise smooth surface; it is the union of the three smooth parametrized
surfaces Si, 52, and 53 described next. We calculate ffs z dS.
The three smooth pieces may be parametrized as follows:
5i (lateral cylindrical surface):
52 (bottom disk):
3 cos s
3 sins
x =
y =
z = 0
s cos?
s sin?
0<s<2n, 0 < / < 15,
0 < s < 3, 0 < t < 2n,
and
53 (top disk):
x = s cos/
v = ssin? 0 < s < 3, 0 < t < lit.
z= 15
Using Definition 2.1, we have
p p pl5 p2tt
I I zdS= I I 1 1|(— 3 sins i + 3 coss j) x k|| ds dt
J Js, Jo Jo
10 JO
■15 p2n
Jo Jo
1 1| 3 sins j + 3 coss i|| ds dt
/i 15 pin pi
= / Itdsdt =
Jo Jo Jo
15
biztdt = 3jrr L = 675n
Now, / fs z dS = 0, since z vanishes along the bottom of 5. For 53, we have
/ / z.dS= [ [ l5dS= 15- / /
J Js, J Js, J Js,
l5dS= 15- / / IdS
is, J Js,
15 • area of disk = 15 • (9n) = \35n.
Therefore,
f f zdS= f f zdS+ [ [ zdS+ f f
J J s J Jsi J Js? J Js,
zdS
' S\ J J Si J J s,
= 6757T + 0 + 1357T = 8107T.
If a surface 5 is given by the graph of z = g(x, y), where g is of class C1 on
some region D in R2, then 5 is parametrized by X(jf, y) = (x, y, g(x, y)) with
(x, y) e D. (See Example 4 of §7.1.) Then, from Example 13 in §7.1,
so that
N(x,y)= -g.Ti-gvj + k,
jjjdS= j j f(x, y, g(x, y))Jgl + g2y+ldx dy.
(4)
7.2 | Surface Integrals 473
; „
D
Figure 7.16 The graph
of 4 — x2 — y2 over the
disk D of radius 2.
Figure 7.17 If u = tan 1 4,
then sec u = y/VJ,
EXAMPLE 3 Suppose S is the graph of the portion of the paraboloid z =
4 — x2 — y2, where (x, y) varies throughout the disk
D = {(x,y) eR \ x + y < 4}.
(See Figure 7.16.) Formula (4) makes it straightforward although rather involved
to calculate
jj (4 - z)dS, where X(x, y) = (x, y, 4 - x2 - y2).
In particular, we have
jj(4-z)dS = jj (4 - (4 - x2 - v2)) 74jc2 + 4y2 + 1 dx dy
= jj (x2 + v2V4jc2 + 4v2 + 1 dx dy.
To integrate, we switch to polar coordinates; that is, we let x = r cos 9 and y =
r sin#, where 0 < r < 2, 0 < 0 < 2n. The desired integral becomes
n7.7T nT. nA n ATT
/ / r2^4r2 +\rdrd6 = / / r^Ar2 +\d9dr
Jo Jo Jo Jo
= 2n / rV4r2+ \dr,
Jo
by Fubini's theorem. Now let 2r = tanw; that is, let r = \ tanw so that dr
-2 P2iT
| sec2 udu. The previous integral transforms into
2;r
I
tan-1 4
tan
3 u • y/ tan2 u + 1 • - sec2 u du
8 Jo
= ~\
8 Jo
= -/
8 Jo
tan3 u sec3 u du
tan2 u sec2 u • (sec u tan udu)
(sec2 u — 1) sec2 m sec u tanudu.
Now, let w = sec m so d w = sec w tan « . Hence, when u = 0,w = 1 and when
m = tan-1 4, «; = -^17- (See Figure 7.17.) Thus, the w-integral becomes
71
Vl7
(w — \)w dw
71 r^< *
w )dw
7t (\ 5 1 3
-id iy
5 3
Tt M2 i— 17 r—
— VT7 Vl7 ,
5 3 J \5 3
71.
1 1
391V17 + 1
60
474 Chapter 7 | Surface Integrals and Vector Analysis
Alternatively, we could calculate the integral /„2 r3>/4r2 + 1 dr using integra-
tion by parts with u = r2 (so du = 2rdr) and dv = r^/Ar2 + 1 dr (so v =
_l_(4r2 + ^3/2)
i/
Figure 7.1 8 The
helicoid of
Example 4.
Vector Surface Integrals
Now we develop a means to integrate vector fields along surfaces, beginning with
a definition.
DEFINITION 2.2 LetX: D — ► R3 be a smooth parametrized surface, where
D is a bounded region in the plane, and let F(x, y, z) be a continuous vector
field whose domain includes S = X(£>). Then the vector surface integral
of F along X, denoted ffx F • dS, is
yy f.js= yy fcxcs.o^nc*,
where N(s, ?) = Ts x T,.
As with line integrals, you are cautioned to be careful about notation for surface
integrals. In the vector surface integral ffx F • dS, the differential term should be
considered to be a vector quantity, whereas in the scalar surface integral ffxfdS,
the differential term is a scalar quantity (namely, the differential of surface area).
EXAMPLE 4 Let F = x i + y j + (z - 2y) k. We evaluate ffx F • dS, where X
is the helicoid
X(s, t) = (s cost, s sinf, t), 0 < s < 1, 0 < f < 27T.
The helicoid is shown in Figure 7.18.
We have
K, t, d(y,z) . d(x,z) . 9(x, y)
N(s, t) = i l H k
v y 3(s, f) 9(5, 0 J d(s, t)
sin f s cos f
0 1
= sin t i — cos tj + sk.
Using Definition 2.2, we obtain
-2ir pi
cos t —s sin t
0 1
j +
cos? -i sin(
sin t s cos t
ff¥-dS=f I F(X(s,t))-N(s,t)dsdt
J Jx Jo Jo
= j (s cos t i + s sin t j
Jo Jo
+ (t — 2s sin t ) k) • (sin ? i — cos t j + s k) ds dt
p lit pi n 27T
= / (st - 2s2 smt)ds dt = / (\s2t — \s3 sin?) | ^dt
Jo Jo Jo
-L (
sin
t) dt = (\t2 + \ cos/)!^ = it2.
7.2 | Surface Integrals 475
EXAMPLE 5 Let f(x, y) be a scalar-valued function of class C1 on a bounded
domain D c R2. Suppose S is the surface described as the graph of z = f(x, y);
that is, S = X(D), where X(x, y) = (x, y, f{x, y)). Then
N(x,v)=-/,i-/vj + k,
so that Definition 2.2 becomes
J£ F • dS = JJ F(x, y, f(x, y)) • (-/, i - /, j + k) dx dy. (5)
Formula (5) will prove to be quite useful. ♦
Further Interpretations
As is the case for vector and scalar line integrals, there is a connection between
vector and scalar surface integrals. Suppose X: D -> R3 is a smooth parametrized
surface and F is continuous on S = X(D). Let N(s, t) = Ts x Tf be the usual
normal vector and let
N(j, t)
n(s, n = .
UNO, on
That is, n is the unit vector pointing in the same direction as N. In particular,
NO, 0 = UNO, Oil Ms, t).
Using Definition 2.2, we have that the vector surface integral is
j j ¥-dS = j j ¥(X(s,t))-N(s,t)dsdt
F(XO, 0) • (UNO, OllnO, 0) ds dt
F(XO,0)-n0,0l|NO,0ll^ dt
i
(F-n)dS. (6)
/x
Since n is a unit vector, the quantity F • n is precisely the component of F in the
direction of n. In other words, formula (6) says that the vector surface integral of F
along X is the scalar surface integral of the component of F normal to S = X(£)).
It is the surface integral analogue of formula (3) of §6.1, which states that the
vector line integral of F along a path x is the scalar line integral of the component
of F tangent to the image curve. To summarize, we have the following results:
//.:
//.:
Line integrals:
.ds= j (F • T)ds.
J X
(V)
Surface integrals:
///'
dS = jj (F-n)dS.
(8)
476 Chapter 7 | Surface Integrals and Vector Analysis
Figure 7.1 9 The amount of fluid transported
across a small piece of S during a brief time
interval At may be approximated by the volume
of a parallelepiped.
As noted in §6. 1 , when x is a closed path, the quantity /x(F • T) ds in equation
(7) is called the circulation of F along x. It measures the tangential flow of F
along the path. On the other hand the quantity f /X(F • n) dS in equation (8) is
known as the flux of F across 5 = X(D). If we think of F as the velocity vector
field of a three-dimensional fluid, then the flux may be thought of as representing
the rate of fluid transported across S per unit time, as we now see. (You may
wish to compare the following discussion with the one in §6.2 concerning the
two-dimensional flux across a curve.)
To avoid notational confusion, we use u and u to denote the parameter vari-
ables for X and t as the variable representing time. Consider a small piece of S,
having area AS, and the amount of fluid transported across it during a brief time
interval At. This amount is the volume determined by F during At. Figure 7.19
suggests that if both AS and At are sufficiently small, then this volume can be
approximated by the volume of an appropriate parallelepiped. Therefore,
Amount of fluid transported « volume of parallelepiped
= (height) (area of base)
= F(X(w0, v0))At ■ n(w0, v0)AS, (9)
since the height of the parallelepiped is the normal component of FAf . We obtain
the average rate of transport across the surface piece during the time interval At
by dividing (9) by Af :
Average rate of transport R» F(X(m0, vo)) • n(w0, vo)AS. (10)
Now, break up the entire surface S = X(Z)) into many such small pieces and sum
the corresponding contributions to the rate of transport in the form given in (10).
If we let all the pieces shrink, then, in the limit as all AS -> 0, we have that the
total average rate AM/ At of fluid transported during Af is approximately
AM f f
Finally, let At —> 0 and define the (instantaneous) rate of fluid transport to be
dM f f
7.2 | Surface Integrals 477
Reparametrization of Surfaces
As seen in §6.1, scalar and vector line integrals over curves depend on the geom-
etry of the curve (and possibly its direction), rather than on the particular way in
which the curve may be parametrized. Much the same is true for surface integrals.
We begin with a definition, analogous to Definition 1.3 of Chapter 6.
DEFINITION 2.3 Let X: D,cR2^ R3 and Y: D2cR2^ R3 be para-
metrized surfaces. We say that Y is a reparametrization of X if there
is a one-one and onto function H: —> D\ with inverse H D\ — >• Z>2
such that Y(s, t) = X(H(j, t)), that is, such that Y = X o H. If X and Y are
smooth and H and H 1 are both of class C1 , then we say that Y is a smooth
reparametrization of X.
EXAMPLE 6 The helicoid parametrized by
X = s cos t
y = s sin t
z = t
0 < s < 1, 0 <t <2tt
may also be described as
x = - cos2f
2
v = - sin2f
2
0 < s < 2, 0 < t < 7t.
z = 2t
The first description corresponds to a map X: [0, 1] x [0, 2tt] — > R3 and the sec-
ond to a map Y: [0, 2] x [0, n] — >• R3. It is not difficult to see that if we make the
change of variables by letting
u = - and v = 2t,
then
Y(s, t) = X(w, v).
Equivalently, we can define a function H: [0, 2] x [0, tt]
H(s, t) = (s/2, 2t). Then H is one-one and onto and Y =
a reparametrization of X.
■> [0, 1] x [0,2jr]with
X o H. Therefore, Y is
♦
EXAMPLE 7 Suppose X is a smooth parametrized surface. Let Y(s, t) =
X(w, v), where u = t, v = s. That is, Y = X o H, where H(s, t) = (t, s). Then
Y is a (smooth) reparametrization that appears to accomplish little. However, if
we let Ny denote the usual normal vector Ts x T, = dY/ds x dY/dt, then we
have
3Y _ aX 9Y _ ax
ds dv dt du '
so that
3Y 9Y 9X 9X 9X 9X
478 Chapter 7 | Surface Integrals and Vector Analysis
The parametrized surface Y is the same as X, except that the standard nor-
mal vector arising from Y points in the opposite direction to the one arising
from X. ♦
The calculation in Example 7 generalizes thus: Suppose X is a smooth
parametrized surface and Y is a smooth reparametrization of X via H, mean-
ing that
Y(s, t) = X(u, v) = X(H(s, /))■
Since H is assumed to be of class C1, we can show from the chain rule that the
standard normal vectors are related by the equation
d(u, v)
ny(^0 = -^-±Nx(u,v). (11)
d(s,t)
(See the addendum at the end of this section for a derivation of formula (11).)
Formula (11) shows that NY is a scalar multiple of Nx. In addition, since H is
invertible and both H and H 1 are of class C1, it follows that the Jacobian of H
is either always positive or always negative. (To see this, note that both H o H 1
and H 1 o H are the identity function. Hence, the chain rule may be applied to
show that the derivative matrix DH(s, t) is invertible for each (s , t); therefore, its
determinant, which is the Jacobian of H, must be nonzero. Since the determinant
is a continuous function of the entries of H, it thus cannot change sign.) Hence,
the standard normal NY either always points in the same direction as Nx or else
always points in the opposite direction (Figure 7.20). Under these assumptions, we
say that both H and Y are orientation-preserving if the Jacobian d(u, v)/d(s, t)
is positive, orientation-reversing if d{u, v)/d(s, t) is negative.
Figure 7.20 If Y is an orientation-reversing reparametrization of X, then NY
points opposite to Nx-
The following result, a close analogue of Theorem 1 .4, Chapter 6, shows that
smooth reparametrization has no effect on the value of a scalar line integral.
THEOREM 2.4 Let X: D\ — » R be a smooth parametrized surface and / any
continuous function whose domain includes X(Dj ). If Y: D2 — »• R3 is any smooth
reparametrization of X, then
fL/ds=fLfds-
7.2 | Surface Integrals 479
The proof of Theorem 2.4 appears in the addendum to this section. With this
result, we can define the scalar surface integral over a smooth surface S by taking
a smooth parametrization X: D —> R3 with S = X(D) that is one-one, except
possibly on 3D. Then we define the scalar surface integral of / on S by
It is a fact (which we shall not prove) that any two smooth parametrizations of
S must be reparametrizations of each other, so Theorem 2.4 tells us that any
particular choice of parametrization we might make does not matter. We need to
assume that X is (nearly) one-one to ensure that the integral is taken only once
over the underlying surface S = X(D). It is also a straightforward matter to extend
these comments to give a definition of a scalar surface integral of a function over
a piecewise smooth surface.
Analogous to Theorem 1.5 of Chapter 6, the following result (whose proof
is in the addendum) tells us that smooth reparametrizations only affect vector
surface integrals by a possible sign change.
THEOREM 2.5 Let X: D\ —> R3 be a smooth parametrized surface and F any
continuous vector field whose domain includes X(Dj). If Y: D2 —> R3 is any
smooth reparametrization of X, then either
if Y is orientation-reversing.
Because of Theorem 2.5, it is a more subtle and involved matter to define
a vector surface integral over a smooth surface than to define a scalar surface
integral. Given a smooth, connected surface, we need to choose an orientation
for it. This is akin to orienting a curve but, perhaps surprisingly, is not always
possible, even for a well-behaved smooth parametrized surface, as Example 8
illustrates.
Here is a formal definition of orientability of a smooth surface.
DEFINITION 2.6 A smooth, connected surface S is orientable (or two-
sided) if it is possible to define a single unit normal vector at each point of
S so that the collection of these normal vectors varies continuously over S.
(In particular, this means that nearby unit normal vectors must point to the
same side of S.) Otherwise, S is called nonorientable (or one-sided).
It is a fact (clearly suggested by Figure 7.21) that a smooth, connected ori-
entable surface S has exactly two orientations.
If S happens to be the image X(D) of a smooth parametrized surface X: D ->
R3, then the normal vectors
if Y is orientation-preserving, or
Ni = Ts x T, and N2 = T, x T.v = -Nj
480 Chapter 7 i Surface Integrals and Vector Analysis
Figure 7.21 The connected orientable surface S shown
with its two possible orientations.
Figure 7.22 Some r-coordinate
curves of the parametrized Mobius
strip.
Figure 7.23 The Mobius strip of
Example 8.
can be used to give unit vector normal vectors ni = Ni/||Ni || andii2 = N2/HN2II
that point in opposite directions. It is tempting to think that ni and 112 always
provide two orientations for S. However, even though both rij and n2 may vary
continuously with respect to the parameters s and t, it is not clear that they must
vary continuously and consistently with respect to the points on the underlying
surface 5. Example 8 is a famous instance of a nonorientable surface.
EXAMPLE 8 The surface parametrized by
-(
1 + t cos - J COS s
2/
1 + t cos sins
0 < s < 2jt,
4 < t < i
2 — — 2
z = t sin
is called a Mobius strip. It may be visualized as follows: The r-coordinate curve
at s = sq is
)S — J t + cos SQ
so"
2
^ t + sin so
k < t < \
2 — — 2
=0
=(■ s
= (™f)<
This is a line segment through the point (cos s0 , sin s0 , 0) and parallel to the vector
s0 cos (y) i + sin s0 cos (y) j + sin (y) k.
COS.'
Several such coordinate curves, marked with the direction of increasing t, are
shown in Figure 7.22. We see that the Mobius strip is generated by a moving line
segment that begins (at s = 0) lying along the positive x-axis, rises to a vertical
position with center at (—1, 0, 0)whens = n, and then falls back to horizontal, but
with direction reversed at s = lit. The s -coordinate curve at t = 0 is parametrized
by
X = coss
y = sins 0 < S < 27T,
z = 0
and so is a circle in the xy-plane. The full Mobius strip is shown in Figure 7.23.
You can make a physical model by taking a strip of paper, giving it a half-twist,
and joining the short ends.
7.2 | Surface Integrals 481
You can understand the gluing process analytically by noting that the map
X:[0,27r]x[-I,I]^R3
defining the Mobius strip as a parametrized surface has the property that X(0, t) =
X(2tt, —t) but is otherwise one-one. Therefore, every point (0, /) on the left edge
of the domain rectangle [0, 2jt] x [— |, |] is mapped to the point (1 + t, 0, 0)of
the Mobius strip, as is the point (2jt, —t) on the right edge of the rectangle. (See
Figure 7.24.)
1/2
-1/2
In
Figure 7.24 Gluing the ends of a strip of paper so that the arrows align
provides a model of the Mobius strip.
Now, let's investigate the orientability of the Mobius strip. The standard nor-
mal vector is
NO, t) = Ts x T, =
We have
and
d(y,z) . d(x,z) . d(x,y)
1 H k
90, t)
90, t)
. s
2
- (cos s + 2t (cos
90, 0
2
cos
t))
+
1
4 cos
4 cos3 - + t (1 +
s / s
cos - ( 1 + 1 cos -
2 V 2
2
coss
cos2 s
N(0,/) = -j
(1 + Ok,
Figure 7.25 Traveling once
around the circular path on the
Mobius strip forces the normal
vector to reverse direction.
N(2;r, -0 = - - j + (1 + 0 k = -N(0, 0-
Therefore, a uniquely determined normal vector has not been defined. More
vividly, imagine traveling along the Mobius strip via the ^-coordinate path at
t = 0, that is, along the circular path
x(j) = X(s, 0) = (coss, sins, 0), 0 < s < 2n.
Follow the standard normal N. At s = 0, it is N(0, 0) = — k, but by the time
we close the loop, it is N(27T, 0) = k. This apparent reversal of the normal
vector means that the strip is not orientable at all — it is one-sided. (See
Figure 7.25.) ♦
A smooth, orientable surface together with an explicit choice of orientation
for it is called an oriented surface. If S is such a smooth oriented surface, then
we define the vector surface integral of F along S by finding a suitable smooth
482 Chapter 7 i Surface Integrals and Vector Analysis
parametrization X of S such that the unit normal vector N(s, f)/||N(s, t)\\ arising
from the parametrization agrees with the choice of orientation normal. We take
the surface integral to be
F • dS.
By Theorem 2.5, if Y is any orientation-preserving reparametrization of X, the
value of //yF' dS is the same as ffx F • dS, so this notion of a surface inte-
gral over the underlying oriented surface S is well-defined. Even though we may
perfectly well calculate ffx F • dS, where X is the parametrized Mobius strip of
Example 8, it does not make sense to consider the surface integral over the under-
lying Mobius strip, since there is no way to orient it. Similarly, the interpretation
of the vector surface integral as the flux of F across the surface only makes sense
once an orientation of the surface is chosen. Then the flux measures the flow rate,
positive or negative, depending on the choice of orientation. (See Figure 7.26.)
Figure 7.27 The sphere
x2 + y2 + z2 = a2 oriented
by outward-pointing unit
normal vectors.
Figure 7.26 How flux depends on orientation. On the
left, the surface S is oriented by unit normal vectors so
that F • n is positive at every point. Hence,
,F-dS =
ffs F • n dS is positive. On the right, S is given the
opposite orientation so that the flux ffsF-dS<0.
Another reason for de-emphasizing the role of parametrization in surface
integrals is that we can often exploit the geometry of the underlying surface and
vector field when making calculations. If S is a smooth, orientable surface and n a
unit normal that gives an orientation of S (so, in particular, n is understood to vary
with the points of S), then, for a continuous vector field F defined on S, we have
dS.
If we can determine a continuously varying, unit normal vector at each point of
S (for example, if S is the graph of a function f(x, y) of two variables or the
graph of a level set f(x, y, z) = c of a function of three variables), then there is
a good chance that the surface integral can be evaluated readily.
EXAMPLE 9 Let F = xi + yj + zkbea radial vector field, and suppose S is
the sphere of radius a with equation x2 + y2 + z2 = a2. Orient S by outward-
pointing unit normal vectors as shown in Figure 7.27. We calculate the flux of
F across S in two ways: (1) by means of a parametrization of S and (2) via
geometric considerations, that is, without resorting to an explicit parametrization
of the sphere.
For approach (1), use the usual parametrization X of the sphere:
x = a cos s sin t
y = a sin s sin t
z = a cos t
0 < s < 2tt, 0 < t < it.
7.2 | Surface Integrals 483
The standard normal vector for this parametrization is
N(s, ?) = — a2 sin? (coss sin? i + sins sin? j + cos? k).
(This normal vector is calculated in Example 11 of §7. 1 .) If we normalize N, we
find that
N(s, 0
n(s, t) = = —(cos s sin? i + sin s sin? j + cos ? k).
I|N(j,OII
Thus, n is inward-pointing at every point on the sphere. Therefore, we must
make a sign change when we evaluate the vector surface integral, if we use the
parametrization just given. Hence, we have
f f F-dS = - f f F-dS = - f f F(X(s,t))-~N(s,t)dsdt
J Js J Jx Jo Jo
n
Jo Jo
2it
(a cos s sin ? i + a sin s sin ? j + a cos ? k)
• (—a2 sin? (cos j sin? i + sins sin? j + cos? k) ds dt
pir p2n
"ill 0 0 0 0 0
= a J I sin? (cos s sin ? + sin s sin ? + cos t)dsdt
Jo Jo
pit pT-TT n7t
= cr" / / sin? ds dt = lira3 I smtdt=4jza3.
Jo Jo Jo
Now, reconsider this calculation along the lines of approach (2). Since S is denned
as a level set of the function f(x, y, z) = x2 + y2 + z2, normal vectors can be
obtained from the gradient:
V(x2 + y2 + z2) = 2x i + 2y j + 2z k.
If we normalize the gradient, then we have unit normal vectors. Thus,
2x i + 2y j + 2z k 2x i + 2y j + 2z k x i + y j + z k
y/Ax1 + 4y2 + 4z2 2v/x2~+y2~+
zr
because x2 + y2 + z2 = a2 at points on S. (Note that n is always outward-
pointing.) Therefore,
JJ.'-M=JJ,r"a
-II.
s
= ff*2 + y2 + *\s=(fa-<ls
J Js a J Js a
= a j j dS = a ■ area of S = a(4na2) = Ancr" . ^
All of the preceding remarks concerning scalar and vector surface integrals
can be adapted to define integrals over piecewise smooth, connected surfaces.
Simply add the contributions of the surface integrals over the various smooth
pieces. The only issue is that of orientation, but assuming that each of the smooth
484 Chapter 7 i Surface Integrals and Vector Analysis
pieces is orientable, then it is possible to provide an orientation to the surface
as a whole. Here's how: Suppose Si and S2 are two smooth surface pieces that
meet along a common edge curve C. Let iij and n2 be the respective unit normal
vectors that give the orientations of Si and S2 ■ Then ni and n2 each give rise
to an orientation of C via a right-hand rule. (To see this, for j = 1, 2, point
the thumb of your right hand along n; ; the direction of your fingers will indicate
the orientation of C.) If C receives opposite orientations from nj and n2, then Si
and S2 are oriented consistently; if C receives the same orientation, then Si and
S2 are oriented inconsistently. (See Figure 7.28.)
Figure 7.28 The piecewise smooth surface S = Si U S2
oriented consistently on the left and inconsistently on the
right.
Figure 7.29 The
piecewise smooth
cylindrical surface S
of Example 10 shown
with orientation
normals.
EXAMPLE 10 We evaluate / fs(x3 i + y3 j) • dS, where S is the closed cylinder
bounded laterally by x2 + y2 = 4, and on bottom and top by the planes z = 0 and
z = 5, oriented by outward normal vectors.
Evidently, S is the union of three smooth oriented pieces: (1) the bottom
surface Si , which is a portion of the plane z = 0, oriented by ni = — k; (2) the top
surface S2, which is a portion of the plane z = 5, oriented by n2 = k; and (3) the
lateral cylindrical surface S3 given by the equation x2 + y2 = 4 and oriented by
normalizing the gradient of x2 + y2 along S3, namely,
2xi + 2y\ xi + y j x'\ + y\
113
74jc2 + 4y2 t/x2^
r
See Figure 7.29 for a depiction of S.
Now we calculate
jjtfi + y'i).dS
j jVi + y3j)-JS +
j ^(x3i+y3j).JS
+ / / (*3i + y3j).JS
i is,
+ y3j)-(-k)JS + JJ^xH + y3])-kdS
+
lLixl+yi)
dS
0 + 0 +
f fsi
\(x* + yA)dS.
7.2 | Surface Integrals 485
To finish the evaluation, we may parametrize 53 by
x = 2 cos s
y = 2sins 0 < s < 2n, 0<t<5.
vz = t
Then
jj(xH+y3j).dS= f f \(x* + yA)dS
s3
= I I j(16cos4j + 16 sin4 2 dt
Jo Jo
= / 16(cos4s + sin4 s)ds dt
Jo Jo
= / / 16((cos2s)2 + (sin2s)2)ds^
Jo Jo
5 r2n / / j + cos2s\2 /l-COs2sx2N
//
Jo Jo
16 + ds dt,
from the half-angle substitution. Thus,
*5 sin
f f(xH + y3\)-dS= f f f(2 + 2 cos2 2s) ds dt
J Js Jo Jo
/>5 rln
= (8 + 4(1 + cos 4s)) ds dt.
Jo Jo
By once again using the half-angle substitution, we get
■5 pS
f f(xH + y3i)-dS= f (12i + sin4s)|2* dt = f 24n dt = 120tt. ♦
J Js Jo Jo
Summary: Surface Integral Formulas
Scalar surface integrals:
For a surface S parametrized by X: D c R2 R3,
[ [ fdS = [ f fdS= f f f(X(s,t))\\TsxTt\\dsdt.
J Js J Jx J Jd
Surface area element is dS = \\TS x Tr|| ds dt.
For a surface S described as a graph of a function z = g(x, y), where g: D c
R2 R,
ff fdS = j j fix, y, g(x, y)) Jgx(x, y)2 + gy(x, yf + 1 dx dy.
Surface area element is dS = ^/ gx{x, y)2 + gy(x, y)2 + 1 dx dy.
486 Chapter 7 i Surface Integrals and Vector Analysis
Vector surface integrals:
For a surface S parametrized by X: D C R2 — >• R3,
jj^F-dS = jf(F-n)dS = j j F(X(s, t)) -~N(s, t)ds dt,
where N = T, x T, and n = N/||N||.
Vector surface integral element is rfS = N(j, t)ds dt.
For a surface 5 described as a graph of a function z = g(x, y), where g: D Q
Rz -> R,
Here n = (-fo i - gy j + k)/^2 + g2 + 1.
Vector surface integral element is S = (— gx i — gy j + k)dx dy.
Addendum: Proofs of Theorems 2.4 and 2.5 —
We begin by establishing formula (1 1) of this section.
■ Lemma Suppose X: D\ -> R3 is a smooth parametrized surface and Y: D2 —>
R3 is a smooth reparametrization of X via H: D2 — > D\, where we denote H(s, f)
by (u, v). Then the standard normal vectors Nx and NY are related by the equation
d(u, v)
NY(M)=!^-fNx(«,i;).
a(s, t)
PROOF First, we set some notation. Since Y is a reparametrization of X via H,
we have, from Definition 2.3, that
Y(s, 0 = X(H(s, 0) = X(ii, u).
(12)
Write (x(s, t), y(s, t), z(s, f)) to denote Y(s, t) and (x(u, v), y(u, v), z(u, v)) to
denote X(m, u), even though this is a small abuse of notation.
By formula (7) of §7. 1, we have
xt t * ^.z) • 9(*>z) . d(x, y)
d(s, t) d(s, t) 9(i, f)
If we apply the chain rule to equation (12), we obtain
DY(s, t) = DX(u, v)DU(s, t).
Writing out this matrix equation, we get
us ut
Vs V,
xs
X,
Xu Xv
ys
yt
y„ yv
_ zs
Zt _
7.2 | Surface Integrals 487
where xs, xt, etc. denote partial derivatives of the component functions of Y and
xu, xv, etc. are the partial derivatives of the component functions of X. It is a
matter of performing the matrix multiplication to check that
[ xs x, ] = first row of DY(s, t)
= first row of the product DX(u, u)DH(j, t)
= (first row of DX(u, v)) • M(s, t)
— [ Xy Xy ]
Similar results hold for the second and third rows of DY(s , t). We may recombine
these results about rows and establish the following matrix equations:
X,
Xu
Xy
Us
ut
y> .
_ y„
yv _
. Vs
vt _
x,
X,
Xy
Us
ut
zt _
. Zu
Zv _
v, _
and
Us U[
vs v,
ys yt
yu
yv
Us Ut
_ZS Zt _
. Zu
Zv _
Taking determinants, we find that
and
d(x, y)
d(x, y) d(u, v)
3(5,0
d(u, v) d(s, t)
d(x, z)
d(x, z) d(u, v)
3(s,t)
d(u, v) d(s, t)
3(y. z)
d(y,z) d(u, v)
d(s,t) d(u,v)d(s,t)
Thus, returning to the original formula for NY, we find that
„ , .\ d(y>z) d(u, v).
Nv s, t) = 1 ■
' 3(ii, v) 3(s, t)
d(u, v)
d(x, z) 9(w, v) . d(x, v) d(u, w),
J H K
3(m, v) d(s, t) d(u, v) d(s, t)
d(s, t)
Nx(m, v),
as desired.
Proof of Theorem 2.4 We use Definition 2.1 and the change of variables
theorem for double integrals. Thus, by Definition 2.1 and the lemma just proved,
fifds
J £ f(Y(s,ty)\\NY(s,t)\\dsdt
f Id2
f(X(H(s, 0))
3(w, u)
3(5, 0
|Nx(m(s, 0. v(s, 0)11 ^Jfi??-
488 Chapter 7 i Surface Integrals and Vector Analysis
From the change of variables theorem, it follows that
I I fdS= II f(X(u,v))\\m.u,v)\\dudv= II fdS,
J Jy J J Di J Jx
by Definition 2.1. ■
Proof of Theorem 2.5 This result can be established along the lines of the pre-
vious proof. Beginning with Definition 2.2 and using the lemma just established,
we have
jj F-dS = jj F(Y(s,t))-~NY(s,t)dsdt
(L
F(X(HO, 0)) • ^4 Nx("(^ 0, v(s, t))ds dt.
D2 90, 0
ds dt,
Therefore,
JJ^F.dS = ±jjD F(X(H(M)))-NX(«(M),KM))
where we take the "+" sign if Y is an orientation-preserving reparametrization
of X (since the Jacobian d(u, v)/d(s, t) is positive and hence equal to its absolute
value) and the "— " sign if Y is orientation-reversing. By the change of variables
theorem, this last expression is equal to
±11 F(X(m, w))-Nx(m, v)du dv = ± 1 1 F-dS,
J J Di J JX
by Definition 2.2. ■
7.2 Exercises
1. Let X(s,t) = (s,s + t,t), 0<s<l, 0<t<2.
Find
(x2 + y2+z2)dS.
2. Let D = {(s, t) \ s2 + t2 < 1, s > 0, t > 0} and let
X: D R3 be defined by X(s, f) = (s + t,s- t, st).
(a) Determine ffxf dS, where f(x, y, z) = 4.
(b) Find the value of ffx F • dS, where F = xi + yj +
zk.
3. Find the flux ofF = xi + 3'j-|-zk across the sur-
face S consisting of the triangular region of the plane
2x — 2y + z = 2 that is cut out by the coordinate
planes. Use an upward-pointing normal to orient S.
4. This problem concerns the two surfaces given para-
metrically as
X(s, t) = (s cos t, s sinr, 3s2),
0 < s < 2, 0 < t < In.
and
Y(s,t) = (2s cost, 2s sin t , 12s2),
0 < s < 1, 0 < t < An.
(a) Show that the images of X and Y are the same.
(Hint: Give equations in x, y, and z for the sur-
faces in R3 parametrized by X and Y.)
(b) Calculate / fx(y i - x j + z2 k) • dS and / fY(y i -
x j + z2 k) • dS. Reconcile your answers.
5. Find ffsx2dS, where S is the surface of the cube
[-2,2] x [-2,2] x [-2,2].
6. Find / fs(x2 + y2) dS, where S is the lateral surface of
the cylinder of radius a and height h whose axis is the
z-axis.
7. Let S be a sphere of radius a.
(a) Find Jfs(x2 + y2 + z2)dS.
(b) Use symmetry and part (a) to easily find / fs y2 dS.
7.2 | Exercises 489
8. Let S denote the sphere x2 + y2 + z2 = a2.
(a) Use symmetry considerations to evaluate ffsxdS
without resorting to parametrizing the sphere.
(b) Let F = i + j + k. Use symmetry to determine
ffs F • dS without parametrizing the sphere.
9. Let S denote the surface of the cylinder x2 + y2
—2 < z < 2, and consider the surface integral
4,
(z-x2-y2)dS.
(a) Use an appropriate parametrization of S to calcu-
late the value of the integral.
(b) Now use geometry and symmetry to evaluate the
integral without resorting to a parametrization of
the surface.
In Exercises 10—18, let S denote the closed cylinder with bot-
tom given by z = 0, top given by z = 4, and lateral surface
given by the equation x2 + y2 = 9. Orient S with outward
normals. Determine the indicated scalar and vector surface
integrals.
I j zdS
■ ///
'•//,
11. / / ydS
13. / / xzdS
14. jj(xi + yj)-dS 15. jj^zk-dS
16. j J y3i-dS 17. jj(-yi + xj)-dS
» /pi.*
In Exercises 19-22, find the flux of the given vector field F
across the upper hemisphere x2 + y2 + z2 = a2, z > 0. Orient
the hemisphere with an upward-pointing normal.
19. F = v j 20. F = yi-x\
21. F = —yi + x j - k 22. F = x2i + xy\ + xzk
23. Let S be the parametrized helicoid X(s, t ) =
(s cos t, s sinf, f), with 0 < s < 2,0 < t < 2n. Deter-
mine the flux ofF = yi + ;tj + z3k across S.
24. Let F = 2xi + 2y\ + z2k. Find ffsF-dS, where S
is the portion of the cone x2 + y2 = z2 between the
planes z = —2, and z = 1, oriented with outward-
pointing normal.
25. Find the flux of F = y3z i — xy j + (x + y + z) k
across the portion of the surface z = yex lying over
the unit square [0, 1] x [0, 1] in the. ty-plane, oriented
by upward normal.
26. Let S denote the tetrahedron with vertices (0, 0, 0),
(1, 0, 0), (0, 2, 0), (0, 0, 3) oriented by outward nor-
mal, and let F = x2 i + 4z j + (y — x) k. Find the flux
of F across S.
27. Let S be the funnel-shaped surface defined by x2 +
y
z2 for 1 < z < 9 andx2 + v2 = 1 forO < z < 1.
(a) Sketch S.
(b) Determine outward-pointing unit normal vectors
to S.
(c) Evaluate ffs F • dS, where F = — yi + xj + zk
and S is oriented by outward normals.
28. The glass dome of a futuristic greenhouse is shaped
like the surface z = 8 — 2x2 — 2y2. The greenhouse
has a flat dirt floor at z = 0. Suppose that the tempera-
ture T, at points in and around the greenhouse, varies
as
T(x,y,z) = x2 + y2 + 3(z-2)2.
Then the temperature gives rise to a heat flux density
field H given by H = — k V T . (Here k is a positive con-
stant that depends on the insulating properties of the
particular medium.) Find the total heat flux outward
across the dome and the surface of the ground if k = 1
on the glass and k = 3 on the ground.
29. The surface given by X(s, t) = (x(s, t), y{s, t), z(s, t)),
where
(a + cos - sinf — sin - su^A cos s
V 2 2 /
2 2
a + cos - sin t — sin - sin 2f ^ sin s
s s
sin - sin t + cos - sin 2t
2 2
a is a positive constant, and 0 < s < 2jt, 0 < t < 2jt,
is known as a Klein bottle.
(a) Use a computer to plot this surface for a = 2.
(b) Determine (and describe) the s -coordinate curve
at t = 0.
(c) Calculate the standard normal vector N along the
i-coordinate curve at t = 0(i.e.,findN(.s, 0)).Note
that X(0, 0) = X(2jt, 0). By comparing N(0, 0)
and N(2jt, 0), comment regarding the orientabil-
ity of the Klein bottle. (See Example 8.)
490
Chapter 7 i Surface Integrals and Vector Analysis
7.3 Stokes's and Gauss's Theorems
Here we contemplate two important results: Stokes's theorem, which relates
surface integrals to line integrals, and Gauss's theorem, which relates surface
integrals to triple integrals. Along with Green's theorem, Stokes's and Gauss's
theorems form the core of integral vector analysis and, as explained in the next
section, can be used to establish further results in both mathematics and physics.
Stokes's Theorem
Stokes's theorem equates the surface integral of the curl of a C1 vector field over a
piecewise smooth, orientable surface with the line integral of the vector field along
the boundary curve(s) of the surface. Since both vector line and surface integrals
are examples of oriented integrals (i.e., they depend on the particular orientations
chosen), we must comment on the way in which orientations need to be taken.
DEFINITION 3.1 Let S be a bounded, piecewise smooth, oriented surface
in R3 . Let C be any simple, closed curve lying in S. Consider the unit normal
vector n that indicates the orientation of S at any point inside C. Use n to
orient C" by a right-hand rule, so that if the thumb of your right hand points
along n, then the fingers curl in the direction of the orientation of C. (Equiva-
lently, if you look down the tip of n, the direction of C should be such that the
portion of S bounded by C is on the left.) We say that C with the orientation
just described is oriented consistently with S or that the orientation is the
one induced from that of S. Now suppose the boundary dS of S consists
of finitely many piecewise C 1 , simple, closed curves. Then we say that 3 S
is oriented consistently (or that 95 has its orientation induced from that
of S) if each of its simple, closed pieces is oriented consistently with S.
Some examples of oriented surfaces with consistently oriented boundaries
are shown in Figure 7.30. If the orientation of 5 is reversed, then the orientation of
3 S must also be reversed if it is to remain consistent with the new orientation of S.
Now we state a rather general version of Stokes's theorem, a proof of which
is outlined in the addendum to this section.
Figure 7.30 Examples of oriented surfaces
and curves lying in them having consistent
orientations. On the right, the boundary of S2
consists of three simple, closed curves.
7.3 I Stokes's and Gauss's Theorems 491
THEOREM 3.2 (Stokes's theorem) Let S be a bounded, piecewise smooth,
oriented surface in R3. Suppose that dS consists of finitely many piecewise C1,
simple, closed curves each of which is oriented consistently with S. Let F be a
vector field of class C1 whose domain includes S. Then
V x F-dS = <f> F • ds.
J as
Figure 7.31 The
paraboloid
z = 9 - x2 - y2
oriented with upward
normal n. Note that
the boundary circle
C is oriented
consistently with S.
Theorem 3.2 says that the total (net) "infinitesimal rotation," or swirling, of
a vector field F over a surface S is equal to the circulation of F along just the
boundary of S.
EXAMPLE 1 Let S be the paraboloid z = 9 — x2 — y2 defined over the disk
in the xy-plane of radius 3 (i.e., S is defined for z > 0 only). Then dS consists of
the circle
C = {(x,y,z)\x2 + y2 = 9, z = 0}.
Orient S with the upward-pointing unit normal vector n. (See Figure 7.31.) We
verify Stokes's theorem for the vector field
We calculate
V x F
F = (2z-y)i + (x + z)j + (3x-2y)k.
i j k
d/dx d/dy d/dz
2z — y x + z 3x — 2y
= (-2 - l)i + (2 - 3) j + (1 - (-1)) k = -3i - j + 2k.
An upward-pointing normal vector N is given by
N = 2xi + 2yj + k.
(This vector may, of course, be normalized to give an "orientation normal" n.)
Therefore, using formula (5) of §7.2 we have, where D = {(x, y) \ x2 + y2 < 9},
y^VxF-JS= j j (-3i- j + 2k)-(2xi + 2yj + k)Jx^y
(— 6x — 2y + 2) dx dy
= 11 —6xdxdy— 11 2ydxdy + I I 2dxdy.
J Jd J Jd J Jd
By the symmetry of D and the fact that — 6x and 2y are odd functions, we have
that the first two double integrals are zero. The last double integral gives twice
the area of D. Thus,
VxF-<iS = 2-7r(32)= 18^.
On the other hand, we may parametrize the boundary of S as
x = 3 cos t
y = 3 sin/ 0 < t < 2n.
z = 0
492 Chapter 7 i Surface Integrals and Vector Analysis
Figure 7.32 The surface
S = {(*, y, z) |
Z = e
-(*2+r)
Z > l/e)
has boundary
dS={(x,y,z)\
x2+y2 = l,z= l/e}.
(This parametrization yields the orientation desired for dS.) Then
(f F-ds= f
Jss Jo
F(x(t))-x'(t)dt
-2n
= f (0 - 3 sin?, 3cosf + 0, 9cosf - 6sinf)- (-3 sin/, 3cosf, 0)dt
Jo
*2n
f
= / (9sin2f + 9cos2f)^
Jo
pin
9dt = 18tt,
which checks.
EXAMPLE 2 Consider the surface S defined by the equation z = e (x2+y2) for
z > l/e (i.e., S is the graph of f(x, y) = e~(A'2+-y2) denned over D = {(x, y) |
x2 + y2 < 1}). Let
F = (ey+z -2y)i+ (xey+z + y) j + ex+-Y k.
Then, no matter which way we orient S, we can see that JL V x F • dS looks im-
possible to calculate. Indeed, suppose we take the upward-pointing normal vector
N = 2xe-(x2+y2) i + 2ye-(x2+y2) j + k.
Then, because
V x F = (ex+y - xey+z) i + (ey+z - ex+y)\ + 2k
(you may wish to check this), using formula (5) of §7.2, we find that
//sV*F.„s
= jj 2xe-{x2+y2)(ex+y - xey+z) + 2ye~^2+y2\ey+z - ex+y) + 2] dx dy.
We will not attempt to proceed any further with this calculation.
It is tempting to use Stokes's theorem at this point, since the boundary of S is
the circle x2 + y2 = 1, z = l/e. (See Figure 7. 32.) If we parametrize this circle by
X = cosf
V = sinf
1
0 < t < 2jt,
z =
then
F • ds
-L
-I
It,
{e
sin t-\-\je
2 sin t , cos t esin '+ 1 /e + sin t , ecos ,+sin ') • (- sin t , cos t , 0) dt
It,
(2 sin2 t - sint esint+1/e + cos2 tesint+1/e + cosf sinf)^-
Again, we have difficulties.
However, the power of Stokes's theorem is that if S' is any orientable
piecewise smooth surface whose boundary dS' is the same as 35 then, subject to
7.3 I Stokes's and Gauss's Theorems 493
5'
n dS = dS'
Figure 7.33 Both S and 5"
have the same boundary and
are oriented as indicated.
Therefore, by Stokes's theorem,
ffsVx*.dS =
fL V x F • dS.
orienting S' appropriately,
F • ds = (b F-ds = V x F • dS.
IS JdS JdS1 J JS'
Hence, we may evaluate ffs V x F • dS by using a different surface! (See Fig-
ure 7.33.)
To use this fact to our advantage, note that V x F has a particularly simple
k-component. Thus, we let S' be the unit disk at z — l/e:
S' = {(x,y,z) | x2 + y2 < l,z = l/e}.
Consequently, if we orient 5" by the unit normal vector n = +k, we have
//IVxF-rfS = //,
-//,
-//.
V x F • dS
(V x F-n)dS
2 dS = 2 • area of S' = 2tx.
Gauss's Theorem
Also known as the divergence theorem, Gauss's theorem relates the vector
surface integral over a closed surface to a triple integral over the solid region
enclosed by the surface. Like Stokes's theorem, Gauss's theorem can assist with
computational issues, although the significance of the result extends well beyond
matters of calculation.
THEOREM 3.3 (Gauss's theorem) Let D be a bounded solid region in R3
whose boundary 3 D consists of finitely many piecewise smooth, closed orientable
surfaces, each of which is oriented by unit normals that point away from D. (See
Figure 7.34.) Let F be a vector field of class C1 whose domain includes D. Then
•—HI.
V-FdV.
Figure 7.34 A solid region D whose
boundary surfaces are oriented so that
Gauss's theorem applies.
By a closed surface, we mean one without any boundary curves, like a sphere
or a cube. The symbol § is used to indicate a surface integral taken over a closed
surface or surfaces.
494 Chapter 7 i Surface Integrals and Vector Analysis
Si
Figure 7.35 The
solid cylinder D of
Example 3.
Gauss's theorem says that the "total divergence" of a vector field in a bounded
region in space is equal to the flux of the vector field away from the region (i.e.,
the flux across the boundary surface(s)).
EXAMPLE 3 Let F be the radial vector field x i + y j + z k and let D be the
solid cylinder of radius a and height b, located so that axis of the cylinder is the
z-axis and the top and bottom of the cylinder are at z = b and z — 0. (See Fig-
ure 7.35.) We verify Gauss's theorem for this vector field and solid region.
The boundary of D consists of three smooth pieces: (1) the bottom surface Si
that is a portion of the plane z = 0 and oriented by the normal iij = — k, (2) the
top surface S2 that is a portion of the plane z = b and is oriented by the normal
vector 112 = k, and (3) a portion of the lateral cylinder S3 given by the equation
x2 + y2 = a2 and oriented by the unit vector 113 = (x i + y j)/a. (The vector 113
may be obtained by normalizing the gradient of f(x, y, z) = x2 + y2 that defines
S3 as a level set.) Then
F-dS = j j F-JS+ J J F-dS+ I j F-dS
D J J Si J JS2 J Js3
= 11 (xi + yj + zk)-(-k)dS+ / / (x i + y j + z k) • kdS
J Js, J Js,
+
J j^xi + yj + zk). (^±H)
x2 + y2
dS
= \l -zdS+ j j zdS+ j j
J JSi J Js2 J JSi
= 0+ [ f bdS+ f f —dS,
J JS2 J JSi a
since along Si , z is 0; along 52> Z is equal to b; and along S3 , x2 + y2 = a2. Thus,
(ft) F • dS = b • area of S2 + a • area of S3 = bna2 + ailnab) = 3na2b,
JJdD
from familiar geometric formulas.
On the other hand,
V-F= AW+ A(v)+l(z) = 3,
dx ay dz
so that
which can be checked readily.
3 d V = 3 • volume of D = 3na b,
In general, ifF = xi + yj + zk and D is a region to which Gauss's theorem
applies, then
§ F-dS= HI V-FdV= ffl
JJdD J J Jd J J J d
3 dV = 3 • volume of D.
Hence,
(x i + y j + z k) • dS = volume of D.
3D
7.3 I Stokes's and Gauss's Theorems 495
Figure 7.36 The union
of the surfaces S and S'
enclose a solid region D
to which we may apply
Gauss's theorem.
Therefore, we may use surface integrals to calculate volumes in much the same
way that we used Green's theorem to calculate areas of plane regions by means
of suitable line integrals. (See §6.2, especially Examples 2 and 3.)
EXAMPLE 4 Let
F = ey coszi + Vx3 + 1 sinzj + (x2 + y2 + 3)k,
and let S be the graph of
z = (1 - x2 - y2)el-x2-iy2 for z, > 0,
oriented by the upward-pointing unit normal vector. It is not difficult to see that
ffs F • dS is impossible to evaluate directly. However, we will see how Gauss's
theorem provides us with elegant indirect means.
Consider the piecewise smooth, closed surface created by taking the union of
S and S', where S' is the portion of the plane z = 0 enclosed by dS (i.e., the disk
x2 + y2 < 1,^ = 0). Orient S' by the downward-pointing unit normal z, = — k as
shown in Figure 7.36. Note that S U S' forms the boundary of a solid region D
and furthermore, that the orientations chosen enable us to apply Gauss's theorem.
Doing so, we have
f f F-JS+ f f F-dS= E F-dS= HI V-¥dV.
J JS J ' S' JJdD J J J D
Now, it is a simple matter to check that V • F = 0 for all (x, y, z). Therefore, the
triple integral is zero, and we find that
so that
jj^¥-dS + jj^F-dS = 0,
JLF-ds=-fLF-ds-
'S J JS'
In other words, because F is divergenceless, Gauss's theorem allows us to replace
the original surface integral by one that is considerably easier to evaluate. Indeed
we have
F-dS
f j F.(-k) dS
j j (x2 + y2 + 3)dx dy,
where R is the unit disk {(x, y) | x2 + y2 < 1} in the plane. Now, we switch to
polar coordinates to find
1 1 (x2 + y2 + 3)dxdy = f f (r2 + 3)rdrd0
J Jr Jo Jo
= (y+\r2)\i_
Jo
Jo
de
EXAMPLE 5 Consider the vector field
= xi + yj + zk
~ (x2 + yi + z2W
496 Chapter 7 i Surface Integrals and Vector Analysis
(This is an example of an inverse square vector field.) The flux across the sphere
x2 + y2 + z2 = a2 oriented by outward unit normal n is given by §s F • dS =
§s F • n dS. The unit normal to S may be computed as
n =
V(x2 + y2 + z2) 2xi + 2yj+2zk xi + yj + zk
||V(x2 + y2 + z2)|| ^4x2 + 4y2 + 4z2
since x2 + y2 + z2 = a2 on the surface of the sphere. In a similar way, we may
write F(x, y, z) as (x i + y j + z k)/a3 whenever (x, y, z) is a point on the sphere.
Hence,
'xi + yj + zk\ /xi+yj + zkN
F-dS
F-ndS
x2 + y2 + z2
dS
1
-(surface area of S)
, cr } a1
= ^{4na2) = 4jt.
Now we note that the partial derivative of the first component of F with
respect to x is
dFi _ (x2 + y2 + z2)3/2 - 3x2(x2 + y2 + z2)1/2 -2x2 + y2 + z2
3x
Similarly,
(x2 + y2 + z2)3
(x2 + y2 + z2)5/2 '
9f2
3y
2y2 + z2
(x2
z2f/2
and
3F3
+ y2 - 2z2
(x2 + y2 +
z2)5/2-
Thus, V • F = 0, and so any triple integral of V • F must be zero. This would seem
to be at odds with Gauss's theorem. There is no contradiction, however. Note that
the vector field F is not defined at the origin. Therefore, the hypothesis that F
be defined throughout the solid region enclosed by S is not satisfied and, hence,
Gauss's theorem does not apply in this situation. ♦
Figure 7.37 A
sphere of radius a
used to understand
the divergence of a
vector field.
The Meaning of Divergence and Curl
Part of the significance of Stokes's theorem and Gauss's theorem is that they
provide a way to understand the meaning of the divergence and curl of a vector
field apart from the coordinate-based definitions of §3.4. By way of explanation,
we offer the following two results:
PROPOSITION 3.4 Let F be a vector field of class C1 in some neighborhood of
the point P in R3. Let Sa denote the sphere of radius a centered at P, oriented
with outward normal. (See Figure 7.37.) Then
divF(P)= lim (j^ F-dS.
>0+ 47Tfl3
PROOF We have, by Gauss's theorem, that
lim
>o+ 4na!
F-dS= lim
3
fl-»-0+ 47T(33
divFdV,
(1)
7.3 I Stokes's and Gauss's Theorems 497
C„ = dS„
where Da is the solid ball of radius a enclosed by Sa. Next, we use a result
known as the mean value theorem for triple integrals, which states that if /
is a continuous function of three variables and D is a bounded connected solid
region in space, then there is some point Q 6 D such that
///.
f(x,y,z)dV = /(g). volume of D.
D
In our present situation, this result implies that there must be some point Q e Da
such that
///.
divFJV = divF(fi). (volume of Da) = (f na3) divF(g). (2)
Applying formula (2) to formula (1), we have
lim $ F-dS = lim divF(g) = divF(P),
a->o+ Ana} Jfs a-*o+
since, as a -> 0+, the ball Da becomes smaller, "crushing" Q onto P. ■
PROPOSITION 3.5 Let F be a vector field of class C1 in a neighborhood of the
point P in R3. Let Ca be the circle of radius a centered at P situated in the plane
containing P that is perpendicular to the unit vector n. (See Figure 7.38.) Then
the component of curl F(P) in the n-direction is
n-curlF(P)= lim — - <f> F-ds,
a->-o+ no1 JCa
Figure 7.38 A circle of where Ca is oriented by a right-hand rule with respect to n.
radius a centered at P.
PROOF Let Sa denote the disk of radius a in the plane of Ca enclosed by Ca . By
Stokes's theorem,
lim — l— S) F • ds = lim — — f f VxF-JS
£i^o+ 7ta2 Jc a^o+ na2 J Js
.0+
J Jsu
= lim — - / / (VxF.n)iJS. (3)
There is a mean value theorem for surface integrals (similar to the mean value
theorem for triple integrals used in the proof of Proposition 3.4) enabling us to
conclude that there must be some point Q in Sa for which
( V x F • n) dS = (V x F(g) • n)(area of Sa)
= na2(V xF(G)-n), (4)
since Sa is a disk of radius a. Therefore, using equations (3) and (4), we find that
1 f _ . 1
lim — - A F-ds= lim — -(jra2V x F(Q) • n)
a->-0+ na2 JCa a^0+ na2
= lim V x F(Q)-n
= V x F(P) • n,
since Q — >• P as a —> 0+ .
498 Chapter 7 i Surface Integrals and Vector Analysis
Propositions 3.4 and 3.5 justify our claims, made in §3.4, about the intuitive
meanings of the divergence and curl of a vector field. The quantity §s F • dS
used in Proposition 3.4 is the flux of F across the sphere Sa, and so
lim
>0+ 47Tfl3
F • dS
n = k
Figure 7.39 The
configuration needed
to calculate the
k-component of
curl F(xo), using
Proposition 3.5.
is precisely the limit of the flux per unit volume, or the flux density of F at P.
Similarly,
1
lim
a^0+ Ttaz
F • ds
is the limit of the circulation of F along Ca per unit area, or the circulation
density of F at P around n. In particular, Proposition 3.5 shows that curlF(P)
is the vector whose direction maximizes the circulation density of F at P and
whose magnitude is equal to the circulation density around that direction (or else
curlF(ip) is 0 if the circulation density is zero).
In fact, we can turn our approach to divergence and curl completely around
and, instead of defining the divergence and curl by means of coordinates and the
del operator and proving Propositions 3.4 and 3.5, use the surface and line integral
formulas of Propositions 3.4 and 3.5 to define divergence and curl and derive the
coordinate formulations from the limiting integral formulas.
Write F as
M(x, y,z)i + N(x, y, z) j + P(x, y, z) k,
where M, N, and P are functions of class C1 in a neighborhood of the point
xo = (xq, yo, Zo)- We will demonstrate how to recover the coordinate formula
(dP
V x F = —
\dy
dN
~dz
i +
dM
~dz
dP
dx
j +
dN dM\
k
dx dy J
from the formula in Proposition 3.5. (A similar argument can be made to derive
the coordinate formula for the divergence, the details of which we leave to you.)
The idea is to let the unit vector n equal, in turn, i, j, and k in the formula in
Proposition 3.5 and thereby to determine the components of the curl.
First, let n = k, so that Ca is the circle of radius a in the horizontal plane
z = zo, oriented counterclockwise around k. (See Figure 7.39.) Then Ca may be
parametrized by
x = xq + a cos /
y = yo + a sin t
Z = ZO
0<t<2n.
Therefore,
& F-ds= f F(x(t))-x'(t)dt
jc„ Jo
/•2tt
= / 0
JO
(M(x(f))i + N(x(t))j + P(x(t))k)-(-a sinr i + a cos t \)dt
•2ti
= a f (-sin? M(x(t)) + cost N(x(t))) dt.
Jo
7.3 I Stokes's and Gauss's Theorems 499
Next, use Taylor's first-order formula (see §4.1) on M and N , which yields, near
x0 (i.e., for (x, y, z) « (x0, yo, Zo)),
M(x, y, z) « M(xo) + Mx(xo)(x - x0) + Mv(x0)(y - y0) + Mz(x0)(z - zo);
JV(x, y, z) « + W*(x0)(x - x0) + Ny(x0)(y - yo) + N:(x0)(z - Zo).
Along the small circle Ca, we have x — xq = a cos/, y — yo = a sin?, and z —
zo = 0, so that, using the approximations for M and TV, we have
/>2ir
F-ds^a / — sin ? [M(x0) + Mx(x0)a cos? + My(x0)a sin? ] dt
Jo
+
a I i
Jo
cos t [N(xo) + Nx(xo)a cos ? + Ny(xo)a sin ? ] dt
' /
Jo
= — aM(xo) / sin? dt — a MT(xo) / sin? cos? J?
2tt
p27T
' /
Jo
a M,(xo) / sin ?J? + aiV(xo) / cos? J?
Jo
-In
+
nLlt nllt
a2Nx(x0) I cos2 t dt + a2 Ny(x0) I sin? cos? J?. (5)
Jo Jo
This last equality holds because M(x0), MA (x0), etc., do not involve ? and so may
be pulled out of the appropriate integrals. You can check that
piit pin t>
I smt dt = J cos? dt = I
Jo Jo Jo
/ sin2 tdt =
Jo Jo
sin? cost dt = 0,
COS t dt = 71.
Therefore, the approximation in (5) simplifies to
F • ds « -na2My(x0) + jra2Nx(x0).
Now, the error involved in the approximation for j>c F-ds tends to zero as Ca
becomes smaller and smaller. Thus,
k • curl F(x0) = lim — *-r £> F-ds
a->-o+ nal JCa
= lim (-Mv(x0) + /V,(x0))
= Nx(x0) - My(x0).
If we let n = j, so that Ca is the circle parametrized by
x = xi) + a sin ?
y = y0 0 < ? < lit,
z = Zo + a cos t
then a very similar argument to the one just given shows that
j-curlF(x0) = M,(xo)-JPl(x0).
Chapter 7 i Surface Integrals and Vector Analysis
Finally, let n = i, so that Ca is parametrized by
x = x0
y = yo + a cos t 0 < t < 2n,
z = zo + a sin t
to find
i • curlF(xo) = Py(xo) - Nz(x0).
We see that the i-, j-, and k-components of the curl of F are as stated in §3.4.
Addendum: Proofs of Stokes's and Gauss's Theorems
Proof of Stokes's theorem
Step 1. We begin by establishing a very special case of the theorem, namely,
the case where the vector field F = M(x, y, z) i (i.e., F has an i-component only)
and where the surface 5 is the graph of z = f(x, y), where / is of class C1 on a
domain D in the plane that is a type 1 elementary region. (See Figure 7.40.) To
be explicit, the region D is
{(x, y) | y(x) < y < S(x), a <x <b),
where y and S are continuous functions. We assume that S is oriented by the
upward-pointing unit normal.
y
(D
^=5(x).
D
®
®
y = j(x)
a
b
Top view
Figure 7.40 The surface S is the graph of f(x, y) for (x, y) in
the type 1 region D shown at the right.
First, evaluate §as F • ds. The boundary 3S consists of (at most) four smooth
pieces parametrized as follows. (The subscripted curves correspond to the encir-
cled numbers in Figure 7.40.)
C
l •
x = t
v = y(t) a <t < b;
z = fit, y(t))
Co
x = b
V = t y(b) < t < 8(b);
Z = f(b, t)
C3 :
x = t
y = 8(t) a <t <b;
z = fit, S(t))
x = a
y = t y(a) < t < 8(a).
z = f(a, t)
The parametrizations shown for C3 and C4 induce the opposite orientations to
those indicated in Figure 7.40. Therefore,
(f F-ds= j F.Js+/ F-d$- I F-ds- j
JdS JCi JC2 Jc3 JC4
F-ds.
7.3 I Stokes's and Gauss's Theorems 501
Consider the integral over C\ . Since F has only an i-component,
f F-ds= f Mdx + 0dy + 0dz= f M(t, y(t), f(t, y(t)))dt.
Jd JC, Ja
The line integral fc F • ds may be calculated in a manner similar to that for
fc F • ds. In particular, we obtain
f ¥-ds= f M(t,S(t), f(t,8(t)))dt.
JC} J a
For the integral over C2, note that x is held constant. Thus,
/ F • ds = f M dx = 0.
Jc2 Jc2
Likewise, fc F • ds = 0. The result is
<f F.ds= f M(t,y(t),f(t,y(t)))dt-f M(t, 8(t), f(t, S(t)))dt
J dS J a J a
= f [M(x,y(x),f(x,y(x)))-M(x,S(x),f(x,S(x)))]dx. (6)
J a
(In this last equality we've made a change in the variable of integration.)
Now we compare the line integral to the surface integral ffs V x F • dS. For
F = M(x, y,z)i, we have V x F = Mzj — My k, so formula (5) of §7.2 yields
J jvxV.dS = j jjMzi-MyK).(-fxi- fy\+k)dxdy
J a J y(
*b r*W / dM df dM\
ly(x)
The chain rule implies
) dydx.
dz dy dy )
3 dM dM df
— (M(x, y, f(x, y)) = — + — -f.
dy dy dz dy
Thus, using the fundamental theorem of calculus, we have
, , *b ,«(*) a
VxF-dS= / (M(x,y, f(x,y)))dydx
J Js Ja Jy(x) 3y
= f -M(x,y,f(x,y))\yyZSy%dx
J a
= f [-M(x,S(x), f(x,S(x))) + M(x,y(x), f(x,y(x)))]dx,
J a
which agrees with equation (6).
We may readily extend this result to surfaces that are graphs of z = f(x, y),
where the point (x, y) varies through an arbitrary region D in the xy-plane via a
two-stage process: First, establish the result for regions D that may be subdivided
into finitely many elementary regions of type 1 and then apply a limiting argument
similar to the one outlined in the proof of Green's theorem in §6.2.
Chapter 7 | Surface Integrals and Vector Analysis
Figure 7.41 The surface S
is a portion of the plane
x = c. If F = Mix, y, z)i,
then F is always
perpendicular to S, in
particular, to any vector T
tangent to 9S.
Figure 7.42 The surfaces Si
and S2 share the curve C as
part of their boundaries. C
receives one orientation as
part of 9 Si, and the opposite
orientation as part of dSz-
Step 2. Still keeping F = Mi, note that the argument given in Step 1 works
equally well for surfaces of the form y = fix, z) — simply exchange the roles of
y and z throughout Step 1 .
It is also not difficult to see that if S is a portion of the plane x = c (where
c is a constant), then Stokes's theorem for F = Mi holds for this case, too. On
the one hand n = ±i for such a plane (depending on the orientation chosen) and
V x F = Mz j — My k, as we have seen. Hence,
/ / VxF-JS= / / (VxF-n)JS = / / 0
J Js J Js J Js
dS = 0.
On the other hand since F has only an i-component, F is always parallel to
n and thus perpendicular to any tangent vector to S, including vectors tangent
to any boundary curves of S. Therefore, <fas F • ds = 0 also. (See Figure 7.41.)
Now, suppose that S = Si U S2, where Si and S2 are each one of the graph of
z = fix, y), the graph of y = fix, z), or a portion of the plane x = c. Assume
that Si and S2 coincide along part of their boundaries, as shown in Figure 7.42.
The surfaces Si and S2 inherit compatible orientations from S. If C denotes
the common part of 3 Si and 3S2, then we may write 3 Si as C\ U C and 3S2 as
C2 U C, where C\ and C2 are disjoint (except at their endpoints). If 3S, is oriented
consistently with S, for i = 1,2, then note that C will be oriented one way as part
of 3 Si and the opposite way as part of 3S2. From this point, let us agree that C
denotes the curve oriented so as to agree with the orientation of 3 Si.
Now Stokes's theorem with F = Mi holds on both Si and S2; on Si we have
/ / VxF-dS= (p F • ds = I ¥-ds+ j ¥■
whereas on S2 we have
/ / V x¥-dS= (t F • ds = I F-ds - I F
J Js? J as, Jc? Jc
ds,
(7)
(8)
in view of the remarks made regarding the orientation of C. Now, we consider
S = Si U S2, noting that C is not part of 3S. We see that
ffvxF-dS=ff VxF-dS+ff
J JS J JSi J J s2
Using equations (7) and (8) and canceling, we find that
V x F-dS.
/ F-ds+ F-ds
JCi Jc2
F-ds.
us
Thus, Stokes's theorem holds in this case, or, indeed, in the case where S can be
written as a finite union Si U S2 U • • • U S„ of the special surfaces just described.
From a practical point of view, any particular surfaces you are likely to
encounter will be decomposable as finite unions of the special types of surfaces
previously described. However, not all piecewise smooth surfaces are, in fact, of
this form. So to finish a truly general proof of Stokes's theorem when F = Mi,
some further limit arguments are needed, which we omit.
Step 3. Finally, by permuting variables, we essentially repeat Steps 1 and 2 in
the cases where F = N(x, y, z) j or F = P(x, y, z)k. In general, by the additivity
7.3 I Stokes's and Gauss's Theorems 503
of the curl, we have that
V x F • dS
j xF-dS = jjvx(Mi + Nj + Pk)-dS
= j jv x(Mi)-dS + JJvx(Nj)-dS
+ j x(Pk)-dS.
Using the versions of Stokes's theorem just established, we see that
//VxF-dS=(f> Mi-ds+f N]-ds+(b Pk-ds,
J Js JdS JdS Jas
= <b (Mi + N\ + Pk)-ds= f F-ds,
Jas Jas
as desired.
Figure 7.43 The type 1 region
D and its shadow region R in the
xy-plane.
Proof of Gauss's theorem
Step 1. We prove a very special case of Gauss's theorem, namely, the case in
which F = P k (where P(x, y, z) is of class C1 on a domain that includes the
solid region D) and where D is an elementary region of type 1, as in Figure 7.43.
We denote the bottom surface boundary of D by S\ and take it to be given by
the equation z = (p(x, y), where cp is of class C1 and, similarly, we let 52 denote
the top surface boundary and assume it is given by the equation z = is(x, y),
where ^ is also of class C1 . The lateral surface boundary is denoted by S3 ; it may
reduce to a curve or be empty but otherwise is a cylinder over the boundary curve
of a region R in the plane forming the shadow of D.
Orienting 3 D = Si U 52 U S3 with outward-pointing normal vectors, we have
F • dS
f f F-dS+ f f F-dS+ f f
J Jsi J Js7 J Js-t
F-dS.
The orientation normal to S\ should be downward-pointing, hence parallel to
<pxi + <py\ — k, the opposite of the normal vector obtained from the standard
parametrization of Si. Therefore, using formula (5) of §7.2, we have
/ / F-dS= / / P(x,y,(p(x,y))k-((pxi + <py]-k)dxdy
J JS, J J R
-JL
P(x, y, <p{x, y))dx dy.
Similarly, the orientation normal to S2 should be upward-pointing, and so
// F-dS= if P(x,y,f(x,y))-(-irxi-fy}+K)dxdy
J J Si J J R
= j j P(x, y, \j/(x, y))dx dy.
Now the lateral surface S3, if it is nonempty, is a cylinder over a curve in the
xy-plane. Hence, S3 is defined by one or more equations of the form g(x , y) = c.
504 Chapter 7 | Surface Integrals and Vector Analysis
Figure 7.44 The regions D\ and
D2 share the surface S as part of
their boundaries. This common
surface S inherits one orientation
as part of dD\ and the opposite
orientation as part of d Di-
li follows that any normal vector to 53 can have no k-component. Thus,
/ / F-dS = I I (Pk-n3)J5= / / 0dS = 0.
J JSx J Js% J Js%
IS3 J JS3
Putting these results together, we find that
F-dS = f f
D J JR
[P(x, y, f{x, y)) - P(x, y, <p(x, y))]dxdy.
On the other hand, if F = Pk, then V • F = dP/dz, so that, by the funda-
mental theorem of calculus,
r r r r r r^'y) dp
/ / / V-FJV= / / / — dzdxdy
J J Jd J Jr Jip(x.v) 9z
-//.
tp(x,y)
[P(x, y, x[r(x, y)) - P(x, y, cp(x, y))]dxdy.
Therefore, Gauss's theorem holds in this special case.
Step 2. We may repeat Step 1 for F = Mi and D an elementary region of
type 2, and for F = A^j and D of type 3. If D is an elementary region of type 4
(meaning D is simultaneously of types 1,2, and 3), then
F • dS
3D
(Mi + N] + PkWS
i)D
Mi-dS +
• ID
Nj-dS+Qp Pk-dS
3D JJdD
=HLv-M,dv+UL
-ffL
-ffL
! V-NjdV +
V .(Mi+ Nj + Pk)dV
V-FdV.
ffL
V-PkdV
Step 3. Suppose that D = D\ U D2, where each of D\ and D2 is a type 4
elementary region, and that D\ and Do coincide just along part of their boundaries
as shown in Figure 7.44. Let S denote the common part of dD\ and dD2- Then
we write dD\ as Si U S and dD2 as S2 U S, where S\ and S2 are disjoint (except
perhaps along portions of their respective boundaries). If we orient dD\ and dD2
with outward normals so as to satisfy the hypotheses of Gauss's theorem, then S
will be oriented one way as part of dDi and the opposite way as part of dD2. Let
us agree that the symbol S denotes the surface oriented so as to agree with the
orientation of 3 D\ .
Applying Gauss's theorem to both D\ and D2, we obtain
ffL
ffL
V-FdV
S7-FdV
F-dS
SD,
F-dS
8D2
i i Si i is
F • dS
F-dS.
7.3 | Exercises 505
7.3 Exercises
Combining these last two equations, we find that
/ / / V-FdV = iff V -FdV + / / / V-FdV
J J JD J J JDi J J J D2
= f f F-dS + f f F-dS
J J Si J Js2
F-dS,
3D
since 3D = S\ U S2.
Step 4. The result of Step 3 may be extended to regions D that can be
decomposed as a union of finitely many type 4 regions. However, not all regions
to which Gauss's theorem applies meet this criterion. Consequently, to finish a
truly general proof, we once again need an argument using suitable limits of
regions and their integrals, which we omit. ■
In Exercises 1—4, verify Stokes s theorem for the given surface
and vector field.
1. S is defined by x2 + y2 + 5z = 1, z > 0, oriented by
upward normal;
F = xzi + yz) + (x2 + y2)k.
2. S is parametrized by X(s, t) = (s cost, s sinf, f), 0 <
s < 1, 0 < t < jt/2;
F = zi + xj + yk.
3. S is defined by x = yl6 — y1 — z1',
F = xi + yj + zk.
4. S is defined by x2 + y2 + z2 = 4, z < 0, oriented by
downward normal;
F = (2y - z) i + (x + y2 - z) j + (4y - 3x)k.
5. Let S be the "silo surface," that is, 5 is the union of
two smooth surfaces S\ and S2, where Si is defined by
x2 -\- y2 = 9, 0 < z < 8
and S2 is defined by
x2 + y2 + (z - 8)2 = 9, z > 8.
Find ffs V x F • dS, where
F = (jc3 + xz + yz2) i + (xyz3 + y1) j + x2z5 k.
In Exercises 6—9, verify Gauss s theorem for the given three-
dimensional region D and vector field F.
6. F = *i+yj + zk,
D = {(x, y, z) I 0 < z < 9 - x2 - y2}
7. F = (y — x) i + (y — z) j + (x — y) k, D is the unit
cube [0, 1] x [0, 1] x [0, 1]
8. F = x2i+ vj + zk, D = {(x,y,z) | x2 + y2 + 1 <
z < 5}
9. F= -y,i7+yj?+Zk, D = {(x,y,z)\a2<x2 +
jx2 + y2 + z2
y2 + z2< b2}
10. Verify that Stokes's theorem implies Green's theo-
rem. (Hint: In Stokes's theorem take F(x, y, z) =
M(x, y)i+ N(x, y)j; that is, assume F is independent
of z and that its k-component is identically zero.)
1 1 . Let S be the surface defined by y = 10 — x2 — z2 with
y > 1, oriented with rightward-pointing normal. Let
F = (2xyz + 5z) i + ex cos yz j + x2y k.
Determine
V x F • dS.
(Hint: You will need an indirect approach.)
12. Let S be the surface defined as z = 4 — 4x2 — y2
with z > 0 and oriented by a normal with nonnegative
k-component. Let F(x, y, z) = x3i + ey j + zexyk..
Find UjVxF' dS. (Hint: Argue that you can inte-
grate over a different surface.)
13. (a) Show that the path x(f) = (cos?, sin?, sin It) lies
on the surface z = 2xy.
(b) Evaluate
(y + cos x) dx + (sin y + z ) dy + x dz,
where C is the closed curve parametrized and ori-
ented by the path x in part (a).
Chapter 7 I Surface Integrals and Vector Analysis
14. Let S consist of the four sides and the bottom face of
the cube with vertices (±a, ±a, ±a). Orient S with
outward-pointing normals. Evaluate //sVxF-JS,
where F = x2yz3 i + x2y j + xex sin yz k.
1 5. Use Stokes's theorem to find the work done by the vec-
tor field F = (x yz — ex) i — xyz j + (x2yz + sin z) k
on a particle that moves along the line segments from
(0, 0,0), then to (1, 1, 1), then to (0, 0, 2), then back to
(0, 0, 0).
16. Let C be a simple, closed curve that lies in the plane
2x — 3y + 5z = 17. Show that the line integral
(3 cos x + z) dx + (5x — ey)dy — 3 y dz
depends only on the area enclosed by C and its ori-
entation, not on its particular shape or location in the
plane.
1 7. Use Gauss's theorem to find the volume of the solid re-
gion bounded by the paraboloids :
z = 3x2+3y2-16.
y and
18. Let S be defined by z = el~x ~y , z > 1, oriented
by upward normal, and let F = x i + y j + (2 — 2z) k.
Use Gauss's theorem to calculate
F-dS.
19. Give a proof of Stokes's theorem for smooth,
parametrized surfaces S = X(D), where X: D c
R2 — > R3. To make the proof easier, assume that X
is of class C2 and that it is one-one on D (in which
case 3S = X(3D)).
20. Use Gauss's theorem to evaluate
F • dS,
where F = zex i + 3y j + (2 — yz ) k and S is the
union of the five "upper" faces of the unit cube
[0, 1] x [0, 1] x [0, 1]. That is, the z = 0 face is not
part of S. (Hint: Note that S is not closed, so to apply
Gauss's theorem you will have to close it up.)
21. In this problem, let f(x, y, z) be a scalar- valued func-
tion of class C1, and let D be a region in space to
which Gauss's theorem applies. Let n = (m, «3)
be the outward unit normal vector to S = 3D.
(a) If a is any constant vector and F = /a, show that
V-F = V/-a.
(b) Use part (a) with a = i to show that
3/
part
fn\ dS ■
III.
dV. Also obtain similar
S JJJD°*
results by letting a equal j and k.
(c) Define a vector quantity (ft) f dS
fndS
fndSby
fmds, (fh fn2ds, <th fmds
S w S vv s
With notations and definitions as above, show that
if-ds=ffL
VfdV.
(Note that the right side is a triple integral of a
vector-valued expression, so it is also computed
by integrating each scalar component function.)
22. Given a liquid with constant density S, introduce co-
ordinates so that the (flat) surface of the liquid is the
xy-plane and the z-coordinate measures the depth of
the liquid from the surface. (That is, the positive z-axis
points down into the liquid. ) Then the pressure p inside
the fluid due to gravity is given by p(x, y, z) = Sgz,
where g is acceleration due to gravity. Suppose that a
solid object is immersed in the liquid. If the object fills
out a region D in space, then the total buoyant force
on the solid is the total liquid pressure on the boundary
surface S = 3D and is given by
B
pndS,
23.
where n is the outward unit normal to S. (The negative
sign arises because the pressure causes a force point-
ing inward on the object.) Use the previous exercise to
demonstrate Archimedes' principle: The magnitude of
the total buoyant force on an object equals the weight
of the liquid displaced.
Write a careful proof of the three-dimensional case of
Theorem 3.5 of Chapter 6: If F is a vector field of class
C1 whose domain is a simply-connected region R in
R3, then F = V/ for some (scalar- valued) function /
on R if and only if V x F = 0 at all points of R.
24. Let Sr denote the sphere of radius r with center at the
origin, oriented with outward normal. Suppose F is of
class C 1 on all of R3 and is such that
F- dS = ar + b
for some fixed constants a and b.
(a) Compute
///„
V-FdV,
where D = {(x, y, z) | 25 < x2 + y2 + z2 < 49}.
(Your answer should be in terms of a and b.)
(b) Suppose, in the situation just described, that F =
V x G for some vector field G of class C1 . What
conditions does this place on the constants a and i>?
7.3 | Exercises 507
25. Letnfx, y, z) be a unit normal to a surface S. The direc-
tional derivative of a differentiable function f(x, y, z)
in the direction of n is called a normal derivative of
/, denoted df/dn. From Theorem 6.2 of Chapter 2,
we have
9/
V/-n.
(a) Let 5 denote the portion of the sphere x2 + y2 +
z2 = a2 in the first octant (i.e., where x > 0,
y > 0, z > 0), oriented by the unit normal that
points away from the origin. Let f(x, y, z) =
In (x2 + y2 + z2). Evaluate
9/
dn
dS.
(b) Let D denote the piece of the solid ball x2 + y2 +
z2 < a2 in the first octant; that is,
D
[(x,y,z) \x2 + y2 + z2 <a2,
x > 0, y > 0, z > 0}.
Compute fffD V • (Vf)dV, where / is as in
part (a).
(c) Apply Gauss's theorem to the integral in part (b),
and reconcile your result with your answer in
part (a).
26. Suppose that / is such that for any closed, oriented
surface S,
3/
— dS = 0.
dn
(See Exercise 25 for the definition of the normal deriva-
tive df/dn.) Show that then
d2f d2f d2f
— - H H = 0
dx2 dy2 dz2
(i.e., that / is harmonic).
27. Following Proposition 3.4, show that
divF(/>) = lim - <fj)F-dS,
v^o V JTs
where S is a piecewise smooth, orientable, closed sur-
face 5 enclosing a region D of volume V . (Take S to
be oriented by outward normal.) The limiting process
should be assumed to be such that D shrinks down to
the point P.
28. Use the result of Exercise 27 to establish the formula
for the divergence of a C1 vector field
F = Fx(x, y, z)i+ F2(x, y, z)j + F3(x, y, z)k.
That is, show that
dF\ dFi dF-i
divF= — !- + — + —
dx dy dz
in the following manner: Let P have coordinates
(*o, yo, zo) and consider the (small) cube S, of edge
length a, centered at P with faces parallel to the coor-
dinate planes. Note that the volume V enclosed by S
is ai. It will help to recall that if f(x, y, z) is differen-
tiable, then
dx
lim
f(x + Ax,y,z)- f{x,y,z)
Ax
= lim
f(x + ^,y,z)-f{x
Ax
2 '
y. z)
Ax
29. In this problem, you will use the result of Exercise 27 to
find an expression for V • F in cylindrical coordinates.
(See Theorem 4.5 of Chapter 3.) Begin by writing
F = Fr e, + Fe ee + Fz ez ,
where Fr(r, 9, z), Fg{r, 6, z), and Fz(r, 9, z) denote the
components of F in the e,--, eg-, and e; -directions
(respectively). Let P have cylindrical coordinates
(r,9,z). Consider the small "cylindrical coordinate
cuboid" S shown in Figure 7.45. The pairs of oppo-
site faces correspond to values
r - Ar/2 and r + Ar/2;
9-A9/2 and 9 + A9 /2;
z - Az/2 and z + Az/2.
Note that the volume of the cuboid is approximately
r A9 Ar Az.
(a) Approximate §s F • dS (where S is oriented by out-
ward unit normal) by noting that each face of S is
roughly flat with an "obvious" unit normal vector
and that F is approximately constant on each face.
(b) Use your answer in part (a) to calculate the diver-
gence in cylindrical coordinates as
1 (rFr) H .
dz r dr r d9
divF
(This agrees with formula (4) of §3.4.)
Figure 7.45 The cylindrical
coordinate cuboid of
Exercise 29.
Chapter 7 I Surface Integrals and Vector Analysis
30. Use the ideas of Exercises 27 and 29 to calculate the
divergence in spherical coordinates. (See Theorem 4.6
of Chapter 3.) You will want to make use of the small
"spherical coordinate cuboid" S shown in Figure 7.46.
31,
Figure 7.46 The spherical coordinate
cuboid of Exercise 30. The volume of
the cuboid is approximately
p2 simp AO A<p Ap.
The pairs of opposite faces of S correspond to values
p -
Ap/2
and
P + Ap/2;
<p -
Acp/2
and
<P + A<p/2;
e -
AO/2
and
0 + Ad 12.
Let F be a vector field of class C in a neighborhood
of the point P in R3 , and let n be a unit vector drawn
with its tail at P . Let C be a simple, closed curve such
that there is an orientable surface S bounded by C that
contains P and such that n is normal to S at P. Orient
S by using n, and orient C consistently with S. Follow-
ing Proposition 3.5, show that, if A denotes the area of
S, then
n-curlF(P)
1
lim —
A^O A
F • ds.
Here the limiting process is assumed to be such that C
shrinks down to the point P. (See Figure 7.47.)
(a) Find the ez-component of curlF by considering
the planar path shown in Figure 7.48. The pairs of
opposite "edges" of the approximately rectangular
Figure 7.48 The path C of
Exercise 32(a).
path C correspond to the values
r — Ar/2 and r + Ar/2,
and
0-A0/2 and 6 + A6/2
(all with constant z-coordinate). Note that the area
enclosed by C is approximately r AO Ar. Approx-
imate the line integral fc F • ds by using the fact
that, for small AO and Ar, each edge of C is
roughly straight. Show that
1 dFr 13
• curl F :
+ (rFg).
r d0 r or
(b) Use the path in Figure 7.49 to show that
1 dFz dFe
r~d0 dz'
er • curl F :
Figure 7.47 Figure for
Exercise 3 1 .
32. In this problem, you will use the result of Exercise 3 1
to determine an expression for curl F in cylindrical co-
ordinates. Begin by writing
Figure 7.49 The path C of
Exercise 32(b).
(c) Use the path in Figure 7.50 to show that
dFr dFz
dz dr
Combine this with the results of parts (a) and (b)
to obtain
eg • curl F =
curlF =
1
er
ree ez
d/dr
d/90 d/dz
Fr
rFg Fz
F = Fr er + Fe e9 + Fz ez
(See Theorem 4.5 of Chapter 3.)
7.3 | Exercises
Az
Ay-
Figure 7.50 The path C of
Exercise 32(c).
33. In this problem, you will determine an expression for
curl F in spherical coordinates. Let F be a vector field
of class C1, and write
F = Fpep + Fve(p + F0ee.
Show that
1
ep • curl F
p smcp
9 9F„
— (sinpFfl)- — —
dip 86
and
e9 • curl F
eg • curl F
1 dFp 9
sinip 90 9p
dp
(pF<p) "
3f„
9^
by using Exercise 3 1 and the three paths shown in Fig-
ure 7.51. Conclude that
curl F
1
pL sin <p
p sin <p eg
9/9p
d/d(p
d/de
Fp
pF<p
p sin (p Fe
(See Theorem 4.6 of Chapter 3.)
34. Of the six planar vector fields shown in Figure 7.52,
four have zero divergence in the regions indicated and
three have zero curl. By considering appropriate in-
tegrals and using the results of Exercises 27 and 3 1 ,
categorize each vector field.
Figure 7.51 The paths of Exercise 33.
;•!»!»;•
\ 1 1' i ,
(b)
(d)
Figure 7.52 Figures for Exercise 34.
Chapter 7 i Surface Integrals and Vector Analysis
7.4 Further Vector Analysis; Maxwell's
Equations
In this section, we use Gauss's theorem and Stokes's theorem first to prove some
abstract results in vector analysis and then to study Maxwell's equations of elec-
tricity and magnetism.
Green's Formulas
Our purpose in §7.4 is to establish a few fundamental results of vector analysis.
Throughout the discussion all scalar and vector fields are defined on subsets of R3 .
The following pair of results is established readily:
THEOREM 4.1 (Green's first and second formulas) Let / and g be scalar
fields of class C2, and let D be a solid region in space, bounded by a piecewise
smooth surface S = dD, oriented as in Gauss's theorem. Then we have
Green's first formula:
/ / fDVf-V8dV + / / fDfV2gdV = §sfV8'dS-
Green's second formula:
III
(fV2g-gV2f)dV = (fb(fVg-gVf) -dS.
D MS
PROOF The first formula follows from Gauss's theorem applied to the vector
field F = /Vg. (We leave the details to you.) The second formula follows from
writing the first formula twice:
/ fLvf'v§dv+ / / Lfv28dv= §sfV8'ds' (1)
/ ILV8'vfdv+ 1 1 L8V2fdv=§s8Vf'ds' (2)
Now, subtract equation (2) from equation (1). ■
The third of Green's formulas requires considerably more work to prove.
THEOREM 4.2 (Green's third formula) Assume / is a function of class C2.
Then, for dD = S oriented as in Gauss's theorem and points r in the interior
ofD,
1 f f f V2 / (x)
4u K\ JW V hi- — xii y mi- — xii 7
In this formula, d V denotes integration with respect to the variables in x =
(x, y, z) (i.e., r = (ri, r2, r3) is fixed in the integration), and the symbol V means
Vx, differentiation with respect to x, y, and z.
A proof of Theorem 4.2 appears in the addendum to this section.
7.4 | Further Vector Analysis; Maxwell's Equations
An Inversion Formula for the Laplacian
Green's third formula is a type of inversion formula — a formula that enables
us to recover the values of a function / by knowing certain integrals involving
its gradient and Laplacian. Green's third formula is mainly of technical interest
in proving further results. We use it here to obtain an inversion formula for the
Laplacian operator.
We begin by applying the Laplacian V2 to Green's third formula:
Vr2/(r) = Vr2
1 [[[ Yl
in J J Jd llr-
xJ — dV
An
(3)
The trick is to move V2 inside the surface integral, which is justified since x^r
when x varies over S:
V^t^-/(x)V*fe»»-JS
s \l|r-x|| \||r-x"
By direct calculation, V2(l/||r — x||) = 0 for x ^ r, so
' Vllr-il
Similarly, since /(x) does not involve r,
Vr2 (7(x)Vx f^-*— - H = /(x)Vr2Vx /
= /(x)VxVr2
|r-x|
1
= 0.
Therefore, the Laplacian of the original surface integral is 0. We may conclude
from equation (3) that, for r in the interior of D,
4jt r JJJd ||r-x||
and we have shown the following:
THEOREM 4.3 If <p = V2/ for some function / of class C2, then for r in the
interior of D,
-Lv2 [[[ Jt
to rJJJD\\r
Theorem 4.3 provides an inversion formula for the Laplacian in the following
sense: If V2/ = <p, then
where g is any harmonic function (i.e., g is such that V2g = 0). That is, if the
Laplacian of / is known, we can recover the function / itself, up to addition of
a harmonic function.
Chapter 7 | Surface Integrals and Vector Analysis
Figure 7.53 The sphere Sj,
of radius b, centered at the
origin, together with an
outward unit normal vector.
Finally, it can be shown that the result of Theorem 4.3 holds under consider-
ably less stringent hypotheses than having <p be the Laplacian of another function. 1
Maxwell's Equations
Maxwell's equations are fundamental results that govern the behavior of — and
interactions between — electric and magnetic fields. We see how Maxwell's equa-
tions arise from a few simple physical principles coupled with the vector analysis
discussed previously.
Gauss's law for electric fields If E is an electric field then the flux of E
across a closed surface S is
Flux of E = (fl)E-dS.
(4)
(5)
Applying Gauss's theorem to formula (4), we find that
Flux of E = jjj V-EdV,
where D is the region enclosed by S.
If the electric field E is determined by a single point charge of q coulombs
located at the origin, then E is given by
q x
E(x)
(6)
4jt€0 ||X||
where x = x i + y j + z k. In mks units, E is measured in volts/meter. The constant
€q is known as the permittivity of free space; its value (in mks units) is
8.854 x 10-12 coulomb2 /newton-meter2.
For the electric field in equation (6), we can readily verify that V • E = 0 wherever
E is defined. From formulas (4) and (5), if S is any surface that does not enclose
the origin, then the flux of E across S is zero.
But now a question arises: How do we calculate the flux of the electric field
in equation (6) across surfaces that do enclose the origin? The trick is to find
an appropriate way to exclude the origin from consideration. To that end first
suppose that Sb denotes a sphere of radius b centered at the origin (i.e., Si, has
equation x2 + y2 + z2 = b2). Then the outward unit normal to Si, is
xi + yj + zk 1
= - x.
(See Figure 7.53.) From equation (6),
On Sb, we have ||x||
E-dS =
Si,
b, so that
E-dS =
Si,
4jre0
x
b~l'
-xdS.
Si,
Si,
-dS= —!—
b 4tt€0
Si,
ixr
b4
dS
4jTe0b2
47T€0b2
_ 1
¥ ~ Aneob2
(surface area of Sb)
dS
Si,
(4jvb2)
' See, for example, 0. D. Kellogg, Foundations of Potential Theory (Springer, 1928; reprinted by Dover
Publications, 1954), p. 220.
7.4 | Further Vector Analysis; Maxwell's Equations
Figure 7.54 The solid region D is
the region inside S and outside S/,.
Now, suppose S is any surface enclosing the origin. Let Si, be a small sphere
centered at the origin and contained inside S. Let D be the solid region in R3
between Sb and 5. (See Figure 7.54.) Note that V • E = 0 throughout D, since
D does not contain the origin. (E is still denned as in equation (6).) Orienting
dD = S U Sb with normals that point away from D, we obtain
(ffiE-dS- £ E-dS= £ E • dS = f f f
MS JJsi, JJdD J J J D
V-EdV = 0,
using Gauss's theorem. We conclude that
E • dS
q_
(7)
for any surface that encloses the origin. By modifying equation (6) for E appro-
priately, we can show that formula (7) also holds for any closed surface containing
a single point charge of q coulombs located anywhere. (Note: The arguments just
given hold for any inverse square law vector field F(x) = &x/||x||3, where k is a
constant. See Exercise 1 3 in this section for details.)
We can adapt the arguments just given to accommodate the case of n discrete
point charges. For i = 1 , . . . , n, suppose a point charge of coulombs is located
at position r, . The electric field E for this configuration is
E(x) =
1
47T60
13 •
(8)
For E as given in equation (8), we can calculate that V • E = 0, except at x = r, .
If S is any closed piecewise smooth, outwardly-oriented surface containing the
charges, then we may use Gauss's theorem to find the flux of E across S by taking
n small spheres Si, S%, . . . , Sn, each enclosing a single point charge. (See Fig-
ure 7.55.) If D is the region inside S but outside all the spheres, we have, by
choosing appropriate orientations and using Gauss's theorem,
dffE-JS-V<jffE-JS=(jff E-dS= f f f V-EdV = 0,
MS j_i MS; JJdD J J JD
since V • E = 0 on D. Hence,
E-dS = ^2<jj) E • dS = — 1i
1
= — (total charge enclosed by S).
(9)
Figure 7.55 D is the solid region inside the
surface S and outside the small spheres
Si , S2 Sn, each enclosing a point charge.
514 Chapter 7 i Surface Integrals and Vector Analysis
To establish Gauss's law, consider the case not of an electric field determined
by discrete point charges, but rather of one determined by a continuous charge
distribution given by a charge density p(x). The total charge over a region D in
space is
so that, in place of formula (8), we have an electric field,
E(r)= — f f f p(x) T~XdV. (10)
IntJJjD l|r-x||3
In equation (10), the integration occurs with respect to the variables in x. (Note:
It is not at all obvious that the integral used to define E(r) converges at points
r e D, where p(r) ^ 0, because at such points the triple integral in equation (10)
is improper. See Exercise 20 in this section for an indication of how to deal with
this issue.)
The integral form of Gauss's law, analogous to that of formula (9), is
E E-dS = — [ [ [ pdV, (11)
JJs eo J J Jd
where 5 = 3D. If we apply Gauss's theorem to the left side of formula (1 1), we
find that
[ [ [ V -EdV = — 1 1 1 pdV.
J J Jd to J J Jd
Since the region D is arbitrary, it may be "shrunk to a point." From this, we
conclude that
p
V-E = — .
(12)
Equation (12) is the differential form of Gauss's law.
Magnetic fields A moving charged particle generates a magnetic field. To be
specific, if a point charge of q coulombs is at position ro and is moving with
velocity v, then the magnetic field it induces is
In mks units, B is measured in teslas. The constant /xo is known as the perme-
ability of free space; in mks units
Ho = Ait x 10~7 N/amp2 « 1.257 x 10~6 N/amp2.
In the case of a magnetic field that arises from a continuous, charged medium
(such as electric current moving through a wire), rather than from a single moving
charge, we replace q by a suitable charge density function p and the velocity of
a single particle by the velocity vector field v of the charges. Then we define the
current density field J by
J(x) = p(x)v(x).
(14)
7.4 | Further Vector Analysis; Maxwell's Equations
Figure 7.56 The
total current /
across S is the flux
of the current
density J across S.
In place of formula (13), we use the following definition for the magnetic field
resulting from moving charges in a region D in space:
B(r) = s////<I)vW*
///DJWx
dV
Mo
r-x\
r — x
dV.
(15)
In equation (15), as in equation (10), the integration is with respect to the vari-
ables constituting x. As in equation (10), it is not obvious that the integrals in
equation (15) are convergent if r e D, but, in fact, they are. (See Exercise 21 in
this section.)
Before continuing our calculations, we comment further regarding the current
density field J. The vector field J at a point is such that its magnitude is the current
per unit area at that point, and its direction is that of the current flow. It is not
hard to see then that the total current / across an oriented surface S is given by
the flux of J; that is,
/
"//.
J-dS.
(16)
(See Figure 7.56.)
Returning to the magnetic field B in equation (15), we show that it can be
identified as the curl of another vector field A (to be determined). First, by direct
calculation,
1
|r-x|
Therefore, equation (15) becomes
B(r)
Mo
An
Iff.
J(x) x Vr
dV.
(17)
We rewrite equation (17) using the following standard (and readily verified) iden-
tity, where / is a scalar field and F a vector field (both of class C2):
V x (/F) = (V x F)/ - F x V/.
Formula (18) is equivalent to
FxV/=(VxF)/-Vx (/F).
Therefore,
1 \ . .1
(18)
J(x) x Vr
= (Vr x J(x))
Vr x
J(x)
V, x
J(x)
J(x)
since J(x) is independent of r. Hence,
AttJJJd ||r-x|| 4;r J J Jd
as r does not contain any of the variables of integration. Consequently,
B(r) = V x A(r),
dV,
Chapter 7 | Surface Integrals and Vector Analysis
C
Figure 7.57 The
closed loop C is
oriented so that it
has a right-hand
relationship with
the direction of
current flow it
encloses.
where
A(r)
IM)
1 1 Id II r
J(x)
dV.
(19)
4tt J J J d l|r-x|
Thus, V • B = V • ( V x A) and so, by Theorem 4.4 in Chapter 3, we conclude that
V-B = 0.
(20)
The intuitive content of equation (20) is often expressed by saying that "magnetic
monopoles" do not exist.
The vector field A in equation ( 1 9) furnishes an example of a vector potential
of the field B. (See Exercises 33-38 in the Miscellaneous Exercises for more about
vector potentials.)
Ampere's circuital law If C is a closed loop enclosing a current /, then
Ampere's law says that, up to a constant, the current through the loop is equal to
the circulation of the magnetic field around C. To be precise,
B • ds = fiQl.
(21)
In equation (2 1 ), we assume that C is oriented so that C and / are related by a
right-hand rule, that is, that they are related in the same way that the orientation of
C and the normal to any surface S that C bounds are related in Stokes's theorem.
(See Figure 7.57.)
From equation (16) for the total current, equation (21) may be rewritten as
j> B-ds = /i,0 J j J-dS,
where S is any (piecewise smooth, oriented) surface bounded by C. Applying
Stokes's theorem to the line integral, we obtain
J J V x B • dS = fi0 J J J • dS.
Since the loop C and surface S are arbitrary, we conclude that
V x B = (IqJ.
(22)
Equation (22) is the differential form of Ampere's law in the static case
(i.e., in the case where B and E are constant in time). In the event that the magnetic
and electric fields are time varying, we need to make some modifications. The so-
called equation of continuity, established in Exercise 5 in this section, states that
dp
V-J =
(23)
The difficulty is that if V x B = /zoJ as in equation (22), then equation (23)
implies that
dp
V • (V x B) = V • QioS)
-Mo
dt
7.4 | Further Vector Analysis; Maxwell's Equations
However, assuming B is of class C2, we must have V • (V x B) = 0, even in the
case where p is not constant in time.
The simplest solution to this difficulty is to modify equation (22) by adding
an extra term. From Gauss's law, equation (12), we must have
dp 9E
dt dt
Thus, if we replace J by J + eo(dE/dt) in equation (22), then we can verify that
V • (V x B) = 0. (See Exercise 16 in this section.) Hence, Ampere's law can be
generalized as
V X B = /X0J + tlQ€o
9E
dt'
(24)
Figure 7.58 The
rate of change of
magnetic flux across
S determines the
electromotive force
around the
boundary C.
The term e0(9E/9/)in equation (24), known as the displacement current density,
was first postulated by James Clerk Maxwell in order to generalize Ampere's law
to the nonstatic case. (In this context, the original current density field J is known
as the conduction current density.)
Equation (24) is not the only possible generalization of equation (22), but
it is the simplest one and is consistent with observation. See Exercise 17 in this
section for other ways to generalize equation (22) to the nonstatic case.
Faraday's law of induction Michael Faraday observed empirically that the
change in magnetic flux across a surface S equals the electromotive force around
the boundary C of the surface. This relation can be written as
d1>
dt
E-Js,
(25)
where 4>(f) = ffs B • dS, and C and S are oriented consistently. (See Figure 7.58.)
If we apply Stokes's theorem to the line integral, we find that
V x E • dS.
Since
we have
d<S>
dt
= — f f B dS= f f — -dS,
dt J Js J Js dt
V x E-dS.
is ^ J Js
Because equation (26) holds for arbitrary surfaces, we conclude that
(26)
Chapter 7 | Surface Integrals and Vector Analysis
Summary Equations (12), (20), (24), and (27) together are known as Maxwell's
equations:
V-E =
P
(Gauss's law);
eo
V B =
0
(No magnetic monopoles);
3B
V x E =
(Faraday's law);
V x B =
3E
Mo J + M-oeo —
at
(Ampere's law).
Maxwell's equations allow us to reconstruct the electric and magnetic fields from
the charge and current densities. They are fundamental to the subject of electricity
and magnetism and provide a fitting tribute to the power of the theorems of Stokes
and Gauss.
Addendum: Proof of Theorem 4.2
The most obvious idea is to use Green's second formula with
However, this function fails to be continuous when x = r, so Gauss's theorem
(and hence Green's formula) cannot be applied so readily. Instead we need to
examine the integrals more carefully.
Throughout the discussion that follows, let Sb denote the sphere of radius b
centered at r. First, we establish some subsidiary results.
■ Lemma 1 If h is a continuous function, then
hm$ J^-dS = 0.
b^JJSb Hr-x||
Figure 7.59 Ifx
is any point on the
surface of the
sphere of radius b
centered at r, then
llr — xll = b.
PROOF The average value of h on Sb is defined to be
§Sbh{,)dS i rr
[hU = surface area of 5, = W %
(See Exercise 9 of the Miscellaneous Exercises.) Thus,
h(x)dS.
h{x)dS = Anb2[h\
As x varies over the surface Sb, we have ||r — x|| = b. (See Figure 7.59.) Hence,
lim
h(x)
dS = lim
1
b-*0 .JJSb b
h(x)dS = lim 4nb \h\w„
b->-0
To clarify the variables with respect to which we differentiate, let Vx denote
the del operator with respect to x, y, and z, and Vr the del operator with respect
to r = (n, r2, r3).
7.4 | Further Vector Analysis; Maxwell's Equations
Figure 7.60 The region
D — B denotes the solid D
with a small ball centered at r
removed.
Lemma 2 With h and Sb as in Lemma 1 ,
lim (th h(x)Vx '
Si,
• dS = -4nh(r).
PROOF Let n = (x — r)/||r — x||, the normalization of x — r. Straightforward
calculations yield
1
1
and
n- Vx
]|r - x
for x on Si. Then
lim § /j(x)Vx
b^O
Sb
(x - r) • (x - r)
dS = lim
b^O
1
b5'
s„
h(x)Vx
n\dS
lim
b^O
Si,
h(x)dS
= l™0-b2(4nb2
= - 4nh(r).
(See the proof of Lemma 1.)
Returning to the proof of Green's third formula, we look at a region to which
we can apply Green's second formula, namely, the region D — B, where B is a
small ball of radius b centered at r. (See Figure 7.60.) By Green's second formula
(since l/||r — x|| is not singular on D — B), we have
///.
dV
-If
J Js-sb
/(x)Vx
1
r — x
Vx/(X)
■dS.
(28)
By direct calculation V^l/Hr — x||) = 0, so equation (28) becomes
SSL
Vx2/(x)
dV
s-s„
/(x)Vx
Vx/(x)
dS.
(29)
We may evaluate the right-hand side of equation (29) by replacing the surface
integral over S — St by separate integrals over S and Sb- Now, we take limits as
b ->• 0. By Lemma 1 with h(x) = Vx/(x) • n,
V*/(x)
dS =
Si,
Vx/OO-n
dS -> 0.
Si,
By Lemma 2,
/(x)Vx
Si,
■ dS -4jr/(r).
Chapter 7 | Surface Integrals and Vector Analysis
Since B shrinks to a point as b -> 0, we see that equation (29) becomes
///.
' d i|r-x|| jjs \ \\\r-x\\; \\r ■
from which Green's third formula follows immediately.
7.4 Exercises
1 . Prove Green's first formula, stated in Theorem 4. 1 .
A function g(x, y, z) is said to be harmonic at a point
(xq, yo, Zo) if 8 is of class C2 and satisfies Laplace s equation
V2g
dx2 dy2 dz2
0
on some neighborhood of (xq, yo, zo). We say that g is har-
monic on a closed region DCR! if it is harmonic at all in-
terior points of D (i.e., not necessarily on the boundary of
D). Exercises 2-4 concern some elementary vector analysis of
harmonic functions.
2. Assume that D is closed and bounded and that 3D is a
piecewise smooth surface oriented by outward unit nor-
mal field n. Let dg/dn denote Vg • n. (The term dg/dn
is called the normal derivative of g.) Use Green's first
formula with f(x, y,z) = 1 to show that, if g is har-
monic on D, then
' d-lds = o.
dn
3. Let / be harmonic on a region D that satisfies the
assumptions of Exercise 2.
(a) Show that
///„
Vf-VfdV
3/
f—dS.
6D dn
(b) Suppose f(x,y,z) = 0 for all (x, y, z) G dD.
Use part (a) to show that then we must have
f(x, y, z) = 0 throughout all of D. (Hint: Think
about the sign of V/ • V/.)
4. Let D be a region that satisfies the assumptions of
Exercise 2. Use the result of Exercise 3(b) to show
that if f\ and fi are harmonic on D and fi(x, y, z) =
fi{x, y, z) on 3D, then, in fact, f\ = fi on all of D.
Thus, we see that harmonic functions are determined
by their boundary values on a region. (Hint: Consider
fx ~ fi.)
5. (a) Suppose a fluid of density p(x, y, z, t) flows with
velocity field F(x, y, z, t) in a solid region W in
space enclosed by a smooth surface S. Use Gauss's
theorem to show that, if there are no sources or
sinks,
V • (pF) :
dp
9?
This equation is called the continuity equa-
tion in fluid dynamics. (Hint: The triple integral
fffw lr dV is the rate of fluid flowing into W,
and the flux of pF across S gives the rate of fluid
flowing out of W.)
(b) Use the argument in part (a) to establish the equa-
tion of continuity for current densities given in
equation (23):
dp
' dt '
V-J:
Let T(x, y, z, t) denote the temperature at the point (x, y, z)
of a solid object D at time t. We define the heat flux density H
by H = —kVT. (The constant k is the thermal conductivity.
Note that the symbol V denotes differentiation with respect to
x,y,z, not with respect to t.) The vector field 'H represents the
velocity of heat flow in D. Lt is a fact from physics that the total
heat contained in a solid body D having density p and specific
heat a is
SSL
apTdV.
Hence, the total amount of heat leaving D per unit time is
97*
SSL
op — dV.
d at
(Here we assume that a and p do not depend on 1.) We also
know that the heat flux may be calculated as
ss
n-ds.
Exercises 6-10 concern these notions of temperature, heat, and
heat flux density.
6. Use Gauss's theorem to derive the heat equation,
97'
op
dt
kV2T.
7. In Exercise 6, suppose that k varies with the points in
D; that is, k = k(x, y, z). Show that then we have
op
dT
97
kVzT + Vk-VT.
8. In the heat equation of Exercise 6, suppose that o, p,
and k are all constant and the temperature T of the
solid D does not vary with time. Show that then T
must be harmonic, that is, that V2J = 0 at all points
in the interior of D.
7.4 I Exercises 521
9. (a) If a, p, and k are constant and the temperature T
of the solid D is independent of time, show that
the (net) heat flux of H across the boundary of D
must be zero.
(b) Let D be the solid region between two concentric
spheres of radii 1 and 2. Suppose that the inner
sphere is heated to 120° C and the outer sphere to
20° C. Use the result of part (a) to describe the rate
of heat flow across the spheres.
1 0. Consider the three-dimensional heat equation
V2u
du
a7
(30)
for functions u(x, y, z, t). (Here V2m denotes the
Laplacian32«/3x2 + d2u/dy2 + d2u/dz2.) In this ex-
ercise, show that any solution T(x, y, z, t) to the heat
equation is unique in the following sense: Let D be a
bounded solid region in R3 and suppose that the func-
tions a(x, y, z) and <p(x, y, z, t) are given. Then there
exists a unique solution T(x, y, z, t) to equation (30)
that satisfies the conditions
T(x, y, z, 0) = a(x, y, z) for (x, y, z) € D,
and
(31)
T(x, y, z, t) = 4>(x, y, z, t) for (x, y, z) 6 3D
and t > 0.
To establish uniqueness, let T\ and T2 be two solutions
to equation (30) satisfying the conditions in (31) and
set w = T\ — T2.
(a) Show that w must also satisfy equation (30), plus
the conditions that
w(x, y, z, 0) = 0 for all (x, y, z) 6 D,
and
w(x, y, z,t) = 0 for all (x, y, z) e 3D and t > 0.
(b) For t > 0, define the "energy function"
1
E{t).
y,z,t)YdV.
Use Green's first formula in Theorem 4.1 to show
that E'(t) < 0 (i.e., that E does not increase with
time).
(c) Show that E(t) = 0 for all t > 0. (Hint: Show that
£(0) = 0 and use part (b).)
(d) Show that w(x, y, z, t) = 0 for all t >0 and
(x, y, z) e D, and thereby conclude the unique-
ness of solutions to equation (30) that satisfy the
conditions in (31).
11. Show that Ampere's law and Gauss's law imply the
continuity equation for J. (Note: In the text, we use the
continuity equation to derive Ampere's law.)
1 2. Suppose that E is an electric field, in particular, a vector
field that satisfies the equation V • E = p/eo- A region
D in space is said to be charge-free if p is zero at
all points of D. Describe the charge-free regions of
{x1
)i+ j + (|z3 - 2z) k.
13. By considering the derivation of Gauss's law for elec-
tric fields, show that, for any inverse square vector
field F(x) = fcs/||x||3, the flux of F across a piecewise
smooth, closed, oriented surface S is
' dS if 5 does not enclose the origin,
is [4jik if S encloses the origin.
1 4. Let a point charge of q coulombs be placed at the origin.
Recover the formula
a x
E = — r
47T60 ||x||
by using Gauss's law in the following way:
(a) First, explain that in spherical coordinates, E(x) =
E(x)ep, that is, that E has no components in either
the ev- or ee-direction. Next, note that E(x) may
be written as E(p) — that is, that [|E[| has the same
magnitude at all points on a sphere centered at the
origin.
(b) Show, using Gauss's law and Gauss's theorem, that
Ms
E(p)ep-dS=
s 6o
where S is any smooth, closed surface enclosing
the origin.
(c) Now let S be the sphere of radius a centered at
the origin. Then the outward unit normal n to S at
(p, <p, 6) is ep. Show that
E(p)dS
(d) Use part (c) to show that E(p)
Conclude the result desired.
q/(4jte0p2).
15. (a) Establish the following identity for vector fields F
of class C2:
V x (V x F) = V(V • F) - V2F.
(Note: V2F = (V • V)F.)
(b) In free space (i.e., in the absence of all charges and
currents), use Maxwell's equations to show that E
and B satisfy the wave equation
32,
V2F
9ZF
where k is a constant. What is k in each case?
(c) Use part (a), Faraday's law, and Ampere's law to
show that
V(V-E)-(V-V)E
-Mo
J + eo
3E
97
Chapter 7 | Surface Integrals and Vector Analysis
(d) Show that, in the absence of any charges (i.e., if
P = 0),
V2E
9J
32E
M0 97 +M06°3^-
16. Verify that if the nonstatic version of Ampere's law
(equation (24)) holds, then V • (V x B) = 0.
17. When Maxwell postulated the existence of displace-
ment currents to arrive at a nonstatic version of
Ampere's law, he was simply choosing the simplest
way to correct equation (22) so that it would be consis-
tent with the continuity equation (23). However, other
possibilities are also consistent with the continuity
equation.
(a) Show that in order to have equation (22) valid in
the static case, then, in general, we must have
3Fi
V x B = hqJ +
dt
for some (time-varying) vector field Fi of class C2 .
(b) By taking the divergence of both sides of the equa-
tion in part (a), show that
3F] 9E
V • = uofoV • — •
dt dt
(c) Use part (b) to argue that, from an entirely math-
ematical perspective, Ampere's law can also be
generalized as
9E
V x B = moJ + Mofo— + F2,
at
where F2 is any divergence-free vector field. Since
no one has observed any physical evidence for F2 's
being nonzero, it is assumed to be zero, as in equa-
tion (24).
18. Suppose that J = crE. (This is a version of Ohm's law
that obtains in some electric conductors — here a is a
positive constant known as the conductivity. ) If p = 0,
show that E and B satisfy the so-called telegrapher's
equation,
9F
a2F
V F = moo" — + At06o- ,
of at
1 9. Let E and B be steady-state electric and magnetic fields
(i.e., E and B are constant in time). The Poynting vec-
tor field P = E x B represents radiation flux density.
Use Maxwell's equations to show that, for a smooth,
orientable, closed surface S bounding a solid region D,
E-JdV.
20. Consider the electric field E(r) defined by equation
(10). Note that the integrals in equation (10) are im-
proper in the sense that they become infinite at points
r e D, where p(r) is nonzero. In this exercise, you will
show that, nonetheless, the integrals in equation (10)
converge when D is a bounded region in R3 and p is a
continuous charge density function on D.
(a) Write E(r) in terms of triple integrals for the in-
dividual components. Let r = (r\, r2, r3) and x =
{x, y, z).
(b) Show that if each component of E is written in the
form fffpf(x)dV, then \f(x)
where K is a positive constant.
(c) It follows from part (b) that if
< Ki
IIL
K
dV
converges, so must fffD f(x)dV, Show that
///„
K
dV
converges by considering an iterated integral in
spherical coordinates with origin at r. (Hint: Look
carefully at the integrand in spherical coordinates.)
21. Consider the magnetic field B(r) defined by equation
(15). As was the case with the electric field in equation
(10), it is not obvious that the integrals in (15) converge
at all r e D. Follow the ideas of Exercise 20 to show
that B(r) is, in fact, well-defined at all r, assuming a
continuous current density field J and bounded region
DinR3.
True/False Exercises for Chapter 7
1. The function X: R2 -* R3 given by X(.s, t) = (2s +
3r + 1 , 4.? — t , s + 2t — 7) parametrizes the plane
9x - y - Uz = 107.
2. The function X: R2 -+ R3 given by X(.s, t ) = (s2 +
3t — 1, .s2 + 3, —2s2 + t) parametrizes the plane x —
ly - 3z + 22 = 0.
3. The function X:(-oo, oo) x (-f , f) -> R3 given
by X(.s, t) = (s3 +3tanr - 1, j3 +3, -2s3 +tanr)
parametrizes the plane x — ly — 3z + 22 = 0.
4. The surface X(s, t) = (s2t, st2, st) is smooth.
Miscellaneous Exercises for Chapter 7
5. The area of the portion of the surface z = xexy lying
over the disk of radius 2 centered at the origin is given
by
\ I y/l + e2xy(x4 + x2y2 + 2xy + \)dy dx.
Jo Jo
6. If S is the unit sphere centered at the origin, then
ffsx3dS = 0.
7. If S is the cube with the eight vertices (±1, ±1, ±1),
then ffs(l + x3y)dS = 0.
8. If S denotes the rectangular box with faces given
by the planes x = ±1, y = ±2, z = ±3, then
ffsxyzdS
0.
9. If S denotes the sphere of radius a centered at the
origin, then
jj(z3-z + 2)dS = j j(x-y5 +2)dS.
10. JJS(— yi + xj) • dS = 0, where S is the cylinder
9,0 < z < 5.
X2 + V2
11. Let S denote the closed cylinder with lateral sur-
face given by y2 + z2 = 4, front by x = 7, and back
by x = — 1, and oriented by outward normals. Then
fsxi- dS = 24jt.
12. If S is the portion of the cylinder x2 + y2 = 16,
-2 < z < 7, then //sVx (yi) -dS = 0.
13. ffsF - dS = 6jt, where S is the closed hemisphere
x2 + y2 + z2 = 1, z > 0, together with the surface
jc2 + y2 < 1, z = 0 and F = yz i ■
■ *z i + 3 k.
1 9. If F is a constant vector field, then §s F • d S = 0, where
5 is any piecewise smooth, closed, orientable surface.
20. §s V x F • <iS = 0, where S is any closed, orientable,
smooth surface in R3 and F is of class C1 .
21. Suppose that F is a vector field of class C1 whose do-
main contains the solid region D in R3 and is such that
||F(.v, y, z)\\ < 2 at all points on the boundary surface
S of D. Then
of S.
V • F dV is twice the surface area
22.
23.
24.
25.
26.
27.
28.
29.
14. If S is the level set of a function /(x , y , z) and V / ^ 0,
then the flux of V/ across S is never zero.
15. A smooth surface has at most two orientations.
16. A smooth, connected surface is always orientable.
17. If F is a vector field of class C1 and S is the ellipsoid
x2 + Ay2 + 9z2 = 36, then |/sVxF-dS = 0.
18. 1/jV xF-rfS has the same value for all piecewise
smooth, oriented surfaces S that have the same bound-
ary curve C.
Miscellaneous Exercises for Chapter 7
If S is an orientable, piecewise smooth surface and F
is a vector field of class C1 that is everywhere tangent
to the boundary of S, then ffs V x F • dS = 0.
If S is an orientable, piecewise smooth surface and F
is a vector field of class C1 that is everywhere perpen-
dicular to the boundary of S, then ffs V x F • dS = 0.
If F is tangent to a closed surface S that bounds a solid
region D in R3, then jjjDV -FdV = 0.
Let S be a piecewise smooth, orientable surface and
F a vector field of class C1. Then the flux of F across
S is equal to the circulation of F around the boundary
of S.
Let D be a solid region in R3 and F a vector field of
class C1 . Then the flux of F across the boundary of D
is equal to the integral of the divergence of F over D.
Suppose that / and g are of class C2 and D is a solid
region in R3 with piecewise smooth boundary surface
S that is oriented away from D. If g is harmonic, then
fffDVf-VgdV = §sfVg-dS.
Suppose that / and g are of class C2 and D is a solid
region in R3 with piecewise smooth boundary sur-
face S that is oriented away from D. If / and g are
harmonic, then §s fVg • dS = — §s gV/ • dS.
If V2/ is known, then / is uniquely determined up to
a constant.
30. If S is a closed, orientable surface, then
■dS = 0.
Figure 7.61 shows the plots of six parametrized sur- / /
faces X. Match each parametric description with the X(j' f) = \cost + 0At \cost coss +
correct graph. j \ /
(a) X(s,t) = (t(s2-t2),s,s2-t2) ^ sin? sin.?J,sinr + 0.1r^sinr coss
(b) X(s, t) = (scost, ssinf, s) 1 \ t 0.2t
-= cos t sin s ) , — I — sin s
(c) X(j, t) = ((2 + cos j)cos/, (2 + coss)sinf, r + sins) V5 / 2 V5
(d) X(s, = (sin* cos?, s, sins sinf) (f) \(s, t) = (s cost, s sin?, t)
524 Chapter 7 | Surface Integrals and Vector Analysis
D
Figure 7.61 Figures for Exercise 1.
2. Consider the unit sphere S with equation x2 + y2 +
z2 = 1 . In this problem, you will provide a parametriza-
tion for (almost all of) the sphere that is different from
the one given in Example 2 of §7.1.
(a) First consider the parametrized plane in R3 :
D = {(s, t, 0) ] (s, t) G R2}.
Note that D is just a copy of R2 sitting in R3
as the jcv-plane. For any point (s, t, 0) G D, ar-
gue geometrically that the line through (s, t , 0) and
(0, 0,1) intersects S in a point other than (0, 0, 1).
(See Figure 7.62.)
(b) Now define X: R2 -+ R3 by letting X(s, t) be the
point of intersection of S and the line joining
{s, t, 0) and (0, 0, 1). Write a set of parametric
equations for the line joining (s, t, 0) and (0, 0, 1)
and use it to give a formula for X(s, t).
(c) Show that X(s, t) is a smooth parametrization of
almost all of S. What points of S are not included
in the image of X? For evident geometric reasons,
the map X defined in this problem has an inverse
map from (almost all of) the sphere to the x v-plane,
called stereographic projection of the sphere onto
the plane.
3. (a) Provide a parametrization for the hyperboloidx2 +
y2 — z2 = 1. (Hint: Use the cylindrical coordinates
z and 0 for parameters.)
(b) Modify your answer in part (a) to give a para-
metrization of the hyperboloid
(c) Let jco and yo be such that x2 + y2 = I. Show that
the lines
Figure 7.62 Figure for Exercise 2.
ll(0 = (a(x0 - y0t), b(x0t + y0), ct)
Miscellaneous Exercises for Chapter 7
and
h(t) = (a(y0t + x0), b(y0 - x0t), ct)
lie in the hyperboloid of part (b).
(d) Show that the lines li and I2 of part (c) also lie in
the plane tangent to the hyperboloid at the point
(ax0, by0, 0).
4. Find the surface area of the portion of the hyperboloid
x2 + y2 — z2 = 1 between z = — a and z = a. (See
Exercise 3(a).)
5. (a) Parametrize the ellipsoid
xl z! zl
a2 + b2+ c2
1.
(b) Use your answer in part (a) to set up an integral
for the surface area of the ellipsoid. Do not evaluate
this integral, but verify that it indicates correctly
the surface area in the case that the ellipsoid is a
sphere of radius a.
6. Let y = f(x), a < x < b, be a curve in the xy-plane.
Suppose this curve is revolved around the x-axis to
generate a surface of revolution.
(a) Explain why X(s, t) = (s, f(s) cos t, f(s) sint)
parametrizes the surface so described. (Hint: Con-
sider the r-coordinate curve.)
(b) Verify that the area of the surface is
2n
/ \f(X)\y/l + {f'(x))2dx.
J a
7. Let the curve y = f(x), 0 < a < x < b, be revolved
around the v-axis to generate a surface.
(a) Find a parametrization for the surface.
(b) Verify that the area of the surface is
In \ xjl+(f'(x))2 dx.
J a
8. Let S denote the surface defined by the equation
z = f(x), a < x < b. (The surface is a generalized
cylinder over the curve z = f(x) in the xz-plane.) Let
C denote a piecewise C1, simple, closed curve in the
.icy-plane. Let D denote the region in the xy-plane
bounded by C and assume that every point of D has x-
coordinate between a and b. Let Si denote the portion
of S lying over D.
(a) Show that the portion of S lying over D is
s'(x)dA,
where s(x) = f* ^1 + (f'(t))2 dt ; that is, s is the
arclength function of the curve z = f(x).
(b) Use part (a) to show that the surface area may also
be calculated from the line integral
I
s(x)dy,
where C is oriented counterclockwise,
(c) Compute the surface area of the portion of
x> 1
Z = — + — .
3 4x
1< x < 3,
lying over the rectangle in the xy-plane having ver-
tices (1, ±2), (3, ±2).
Let f be a piecewise smooth surface in R3 and f:X C R3 — >•
R a scalar-valued function whose domain X contains S. Then
the average value of f on S is the quantity
[fU = Sfsfds = SfsfdS
ffsdS area of S
Exercises 9—11 involve the notion of the average value of a
function on a surface.
9. (a) Explain why the definition of the average value
makes sense.
(b) Suppose that the temperature at points on the
sphere x2 + y2 + z2 = 49 is given by T(x, y, z) =
x2 + y2 — 3z. Find the average temperature.
10. Find the average value of f(x, y, z) = x2ez — y2z on
the cylinder x2 + y2 = 4, 0 < z < 3.
1 1 . Find the average value of f(x , y , z) = x2 + y2 — 3 on
the portion of the cone z2 = 4x2 + Ay2, —2 < z < 6.
1 2. A thin film is made in the shape of the helicoid
X(s, t) = (s cost, ssint,t), 0 < s < 1, 0 < t < 4ir.
Suppose that the mass density (per unit area) at each
point (x, y, z) of the film varies as
S(x, y, z) = Jx2 + y2.
Find the total mass of the film.
Let S be a piecewise smooth surface in R3. Suppose that the
mass density at points (x, y, z) of S is S(x , y , z). Using formu-
las analogous to those in §5.6, we define the (first) moments
of S to be
x8(x, y, z)dS,
SL
yS(x, y, z)dS,
and
j j zS(x, y, z)dS.
Exercises 13—16 involve first moments and centers of mass of
surfaces.
1 3. Find the center of mass of the first octant portion of the
sphere
x2 + y2 + z2 = a2,
assuming constant density.
Chapter 7 I Surface Integrals and Vector Analysis
14. Find the center of mass of the piece of the cylinder
x2 + z2=a2, 0<y<a,z>0.
Assume that the density of the surface is constant.
15. Find the center of mass of a sphere of radius a, where
the density S varies as the square of the distance from
the "south pole" of the sphere.
16. Find the center of mass of the cylinder x1 + z2 = a2,
between y = 0 and y = 2, if the density varies as
8 = x2 + y.
Given a piecewise smooth surface S, the moment of inertia I-
of S about the z-axis is defined by the surface integral
I-
j j (x2 + y2)SdS,
where 8(x, y, z) is mass density. The corresponding radius of
gyration about the z-axis is given by
where M
,SdS. Likewise, the moments of inertia of S
about the x- and y-axes are given by, respectively, the surface
fjy+z2)sds
and
I, = j j(x2 + z2)SdS.
(Compare these formulas with the ones in §5.6.) Exercises 17—
19 concern moments of inertia andradii of gyration of surfaces.
1 7. (a) Calculate /- for the surface S cut from the cone
z2 = 4x2 + Ay2
by the planes z = 2 and z = 4. (This surface is
known as a frustum of the cone.) Assume density
is equal to 1 .
(b) Find the radius of gyration r. of the frustum.
(c) Repeat parts (a) and (b), assuming that the density
at a point is proportional to the distance from that
point to the axis of the cone.
18. Let S denote the cylindrical surface with equation
x2 + y2 = a2, where — b < z < b (a, b positive con-
stants). Assume that the density S is constant along
S.
(a) Find the moment of inertia of S about the z-axis.
(b) Find the radius of gyration of S about the z-axis.
19. Let S be as in Exercise 18.
(a) Find the moments of inertia Ix and Iy of S about
the x- and y-axes.
(b) Find the radii of gyration rx and ry .
20. (a) Prove the following mean value theorem for double
integrals: If / and g are continuous on a compact,
connected region D C R2 , then there is some point
P e D such that
J f fgdA = f(P) jj^dA.
(Hint: Consider the ratio JfDfgdA/ JJDgdA
and use the intermediate value theorem.)
(b) Use the result of part (a) to prove the following:
Let S be a smooth surface oriented by unit normal
n and let F be a continuous vector field on S. As-
sume that S may be parametrized by a single map
X: D — > R3 . Then there is some point P e S such
that
F-dS= [F(P)-n(P)](area of S).
21. Let a be a constant vector and C a smooth, simple,
closed curve. Show that j>c a • ds = 0 in two ways:
(a) directly;
(b) by assuming that C is the boundary of a smooth
surface S.
22. Evaluate §c(x2 + z2)dx + y dy + z dz, where C is
the closed curve parametrized by the path x(f) =
(cos t , sin t , cos2 t — sin2 r).
23. Let / and g be functions of class C2, and let S be a
piecewise smooth, orientable surface. Show that
(JVg)-d»= f f(VfxVg).dS.
as J Js
24. Let / and g be functions of class C2, and let S be a
piecewise smooth, orientable surface. Show that
(fVg + gVf)-ds = 0.
(Hint: Use Exercise 23.)
25. Let / be of class C2, and let S be a piecewise smooth,
orientable surface. Show that
Jdi
(/V/)-</s = 0.
26. Let C be a simple, closed, piecewise C1 planar curve
in R3 . That is, C is contained in some plane in R3 . Let
n = ai + foj + ck denote a unit vector normal to the
plane containing C, and let C be oriented by a right-
hand rule with respect to n.
(a) Show that
1 r
- (t (bz — cy)dx + (cx — az)dy + (ay — bx)dz
2 Jc
= area enclosed by C.
(b) Now show that the result of part (a) reduces to
something familiar in the case that C is a curve in
the Jcy-plane. (See Example 2 of §6.2.)
Miscellaneous Exercises for Chapter 7
27. Suppose S is a piecewise smooth, orientable surface
with boundary dS. Use Faraday's law to show that if E
is everywhere perpendicular to dS, then the magnetic
flux induced by E does not vary with time.
28. Let G be a vector field of class C2. Then by Theorem
4.4 in Chapter 3, V • (V x G) = 0. Therefore, Gauss's
theorem implies that
jj(VxG)-dS = jjj V-(V xG)dV = 0.
Then Stokes's theorem yields
0= f f(VxG)^S=l G-ds.
J J S J 95
Hence, all vector fields G of class C2 are conservative.
How can this be?
Recall that we measure an angle in radians as follows: Place
the vertex of the angle at the center O of a circle of radius a,
so that the angle subtends an arc of the circle of length s. Then
the measure of the angle in radians is
s
We can do something similar in the three-dimensional case by
defining a solid angle as the set of rays beginning at the center
O of a sphere of radius a, so that the rays cut out a portion
of the sphere having surface area A. Then the measure of the
solid angle in steradians is
Figure 7.63 A solid angle
measured on a sphere.
Now suppose that S is a smooth, oriented surface and that a
point O in R3 is chosen that is not in S and so that every line
through O intersects S at most once. In this case, we define
the solid angle relative to O subtended by S as the set of rays
beginning at O that pass through a point of S. We measure this
solid angle by calculating
£i(S,0) = JJ^^.dS, (1)
where x denotes the (varying) vector from O to a point P in S
and S is oriented by a normal that points away from O.
Exercises 29—32 develop some ideas regarding solid angles.
29. In this problem we'll see how the measure of the
solid angle given in equation (1) can be related to the
more geometric notion of measuring solid angles using
spheres. To that end, in the situation described above,
construct a sphere Sa of radius a centered at O . Let Sa
denote the intersection of S„ and the solid angle of S
relative to O. (See Figure 7.64.) By applying Gauss's
theorem to the solid region W between S and Sa , show
that
surface area of St
Figure 7.64 A solid angle subtended by a
surface S.
30. If S is parametrized by X: D C R2 -»• R3, and O de-
notes the origin in R3 , we may use equation ( 1 ) to define
the measure of the solid angle subtended by S for very
general surfaces, including ones that may have some
self-intersections. Show that for such a parametrized
surface O) may be calculated as
- X
y
z ~
dx
dy
dz
~ds~
ds
~dl
dx
dy
dz
- ~dt
dt
dt -
31 . Suppose that S is closed, oriented, and forms the bound-
ary of a solid, bounded, simply-connected region W in
Chapter 7 I Surface Integrals and Vector Analysis
R3. Using equation (1) to define the measure of the
solid angle, show that
f ±4jr if S encloses O
38. Suppose that
Q{S, O)
lo
if S does not enclose O.
32. Suppose that S is the circular disk of radius a in the xy-
plane and centered at the origin. Let O be a moving
point along the z-axis and denote it by (0, 0, z). Use
equation (1) to show that — 2jt < f2(S, O) < 2n. In
addition, show that Q(S, O) jumps by An as O passes
through S.
33. Prove the following: Let F be a vector field of class
C1 defined on R3. If V • F = 0, then there is some
vector field G of class C1 such that F = V x G. This
result provides a converse to Theorem 4.4 of Chapter 3 .
(Hint: Let
G(x,y,z)= / t¥(xt, yt, zt) x rdt,
Jo
where r = x i + y j + z k. Show V x G = F. The
identities
V x (A x B) = AV • B - BV • A
+ (B-V)A-(A- V)B,
dt
[t¥(xt, yt, zt)] = (r • VjF(xt, yt, zt)
+ F(xt, yt, zt),
and
d
dt
[t2F(xt, yt, zt)] = t |-^|)F(jef, yt, zt)]
'I
+ F(xt,yt,zt)\
will prove useful. Also, let X = xt, Y = yt, Z = zt
and note that, by the chain rule,
3F _ 9F dX _ 9F
~dx ~ dxJx ~ 'dX'
etc.)
The vector field G defined in Exercise 33 is called a vector
potential for the vector field F. In Exercises 34—36, determine
a vector potential for the given vector field or explain why such
a potential fails to exist.
34. F = 2xi-yj-zk
35. F = xi + yj+zk
36. F = 2>y\ + 2xz\ - 7x2yk
37. The vector potential G identified in Exercise 33 is not
unique. In fact, show that if G is a vector potential for F,
then so is G + V0, where <fr is any scalar-valued func-
tion of class C2 . (This is known as the gauge freedom
in choosing the vector potential.)
GMm
is the gravitational force field defined on R3 —
{(0,0,0)}.
(a) Show that V • F = 0 by direct calculation.
(b) Show that F / V x G for any C1 vector field G
defined on R3 - {(0, 0, 0)} by using Stokes's the-
orem. (Hint: Take a sphere S enclosing the origin
and break it up into the upper and lower hemi-
spheres. Consider ffs F • dS as the sum of the sur-
face integrals over the two hemispheres.)
(c) Why do parts (a) and (b) not contradict the result
of Exercise 33?
In Exercises 39—44 below, you will derive a type of wave equa-
tion and see how Maxwell s equations can be reduced to this
wave equation. Assume that the electric and magnetic fields E
and B are defined on a simply-connected region in R3; also
let the symbol V denote differentiation with respect to x,y,z
(i.e., not with respect to t).
39. Recall that equation (20) in §7.4 was derived by show-
ing that the magnetic field B had a vector potential
A of class C1 (see Exercise 33 above); that is, that
B = V x A. Use Faraday's law (equation (27)) to show
that the vector field E + dA/dt is conservative. Hence,
we may write
9A
E + ^=V/
for an appropriate scalar-valued function f(x, y, z, t).
40. Use the vector potential A for B in Ampere's law (equa-
tion (24) of §7.4) to conclude that
32A
dt2
Ho J + V V • A - iA,0e0
(2)
(Hint: See part (a) of Exercise 15 of §7.4.)
41. Use Gauss's law (equation (12) of §7.4) to show that
(3)
V2/
p 9
60 dt
42. As noted in Exercise 37, the vector potential A is only
unique up to addition of V0, where (p(x, y, z, t) is a
scalar- valued function of class C2 . That is, any vector
field A = A + V0 also works as a vector potential for
B. However, the function / that arises in Exercises 39-
41 will change.
(a) Show that the function / associated to A is related
to / as
/ = / +
dt
Miscellaneous Exercises for Chapter 7
(b) Show that the condition
V • A = (U,06o
9/
3r '
where A = A + V0, is equivalent to the existence
of solutions to the (inhomogeneous) wave equation
d2<p
V • A - ix0eQ
(4)
Given A and / it is possible to solve equation
(4) for <p, so we may assume that the condition
V • A = fio€odf/dt holds.
43. Given the condition V • A = iiQ^df/dt, show that
equations (2) and (3) become
92A
V2A-
V2/ - Mofo
3r2
S2/
9r2
-/x0J;
P_
(5)
(6)
44. Conversely, suppose that A and / satisfy the condition
V • A = fioeodf/dt and equations (5) and (6). Show
that then E = -dA/dt + V/ and B = V x A must
satisfy Maxwell's equations. Hence, solutions to (4)
enable us to define a vector field A and a scalar field
/ from which we may construct E and B that satisfy
Maxwell's equations.
Vector Analysis in
Higher Dimensions
8.1 An Introduction to
Differential Forms
8.2 Manifolds and Integrals of
k-forms
8.3 The Generalized Stokes's
Theorem
True/False Exercises for
Chapter 8
Miscellaneous Exercises for
Chapter 8
Introduction
In this concluding chapter, our goal is to find a way to unify and extend the
three main theorems of vector analysis (namely, the theorems of Green, Gauss,
and Stokes). To accomplish such a task, we need to develop the notion of a
differential form whose integral embraces and generalizes line, surface, and
volume integrals.
8.1 An Introduction to Differential Forms
Throughout this section, U will denote an open set in R", where R" has coor-
dinates (jci, xz, ■ ■ ■ , xn), as usual. Any functions that appear are assumed to be
appropriately differentiable.
Differential Forms
We begin by giving a new name to an old friend. If /: U c R" — >• R is a scalar-
valued function (of class Ck), we will also refer to / as a differential 0-form,
or just a 0-form for short. 0-forms can be added to one another and multiplied
together, as well we know.
The next step is to describe differential 1 -forms. Ultimately, we will see that a
differential 1-form is a generalization of f{x) dx — that is, of something that can
be integrated with respect to a single variable, such as with a line integral. More
precisely, in R", the basic differential 1-forms are denoted dx\, dxz, ■ ■ . , dx„.
A general (differential) 1-form a> is an expression that is built from the basic
1-forms as
to = Fi(xu xn)dx\ + F2(xi, . . . ,xn)dx2 + ■■■ + Fn(xu xn)dx„,
where, for j = 1, . . . , n, Fj is a scalar- valued function (of class Ck) on U c R".
Differential 1-forms can be added to one another, and we can multiply a 0-form
/ and a 1-form to (both defined on U c R") in the obvious way: If
co = F\ dx\ + F2 dx2 H + Fn dxn ,
then
fco = fF\ dx\ + fFz dx2 H h f Fn dx„.
8.1 | An Introduction to Differential Forms
EXAMPLE 1 In R3, let
co = xyz dx + z2 cos y dy + zex dz and >? = (y — z) dx + z2 sin y dy — 2dz-
Then
a> + r] = (xyz + y — z)dx + z2(cos y + sin y) dy + (zex — T)dz.
If f{x, y, z) = xey — z, then
fco = (xey — z)xyz dx + (xey — z)z2 cos y dy + (xey — z.)zexdz. ♦
Thus far, we have described 1 -forms merely as formal expressions in certain
symbols. But 1 -forms can also be thought of as functions. The basic 1 -forms
dx\, ... , dx„ take as argument a vector a = (a\, a%, . . . , an) in R"; the value of
dxt on a is
dxj (a) = a, .
In others words, dx, extracts the tth component of the vector a.
More generally, for each xo e U, the 1-form co gives rise to a combination
cyXo of basic 1 -forms
tt'xo = ^i(xo)^i H r- F„(x0)£/x„;
coXo acts on the vector a e R" as
«Xo(a) = Fi(x0) Jxj(a) + F2(x0)^2(a) H h F„(x0)dx„(a).
EXAMPLE 2 Suppose o> is the 1-form defined on R3 by
co = x2yz dx + y2z dy — 3xyz dz.
If x0 = (1, —2, 5) and a = (a\, a2, 03), then
W(i-2,5)(a) = -10<ijr(a) + 20Jv(a) + 30<iz(a)
= -10«i +20fl2 + 30fl3,
and, if x0 = (3, 4, 6), then
a>(3 4 6)(a) = 216<ix(a) + 96c?v(a) — 216Jz(a)
= 216fli + 96a2 — 216a3.
The notation suggests that a 1-form is a function of the vector a but that this
function varies from point to point as xo changes. Indeed 1 -forms are actually
functions on vector fields. ♦
A basic (differential) 2-form on R" is an expression of the form
dxj A dxj , i, j = 1, . . . , n.
It is also a function that requires two vector arguments a and b, and we evaluate
this function as
dxt A dxj(a, b) =
dxi(a) dxiQa)
dxj(&) dxj(b)
(The determinant represents, up to sign, the area of the parallelogram spanned
by the projections of a and b in the XjXj -plane.) It is not difficult to see that, for
i, j = 1-
Chapter 8 | Vector Analysis in Higher Dimensions
dxj A dxj = —dxj A dxj
(1)
and
dxj A dxi = 0.
(2)
Formula (1) can be established by comparing dxj A dxj(a, b) with dxj A
dx,(a, b). Formula (2) follows from formula (1). Given formulas (1) and (2),
we see that we can generate all the linearly independent, nontrivial basic 2-forms
on R" by listing all possible terms dx{ A dxj, where i and j are integers between
1 and n with i < j:
dx\ A dx2, dx\ A dxi, . . . , dx\ A dxn,
dx2 A dx-i, . . . , dx2 A dxn,
dx„-i A dxn.
To count how many 2-forms are in this list, note that there are n choices for dx{
and n — 1 choices for dxj (so that dxt ^ dxj in view of (2)), and a "correction"
factor of2so as not to count both dxj Adxj anddxj A dxj in light of (1). Hence,
there are n(n — l)/2 independent 2-forms.
Let x = (xi, jt2, . . . , x„). A general (differential) 2-form on U c R" is an
expression
o) = F\2{~iL)dx\ A dx2 + Fii(x)dxi A dx-i + • ■ • + Fn_\n(\)dxn^\ A dxn,
where each Fy is a real-valued function Fy : C/ c R" -> R. The idea here is to
generalize something that can be integrated with respect to two variables — such
as with a surface integral.
EXAMPLE 3 In R3, a general 2-form may be written as
F\(x, y, z)dy Adz + Fi{x, y, z)dz A dx + F^(x, y, z)dx A dy.
The reason for using this somewhat curious ordering of the terms in the sum will,
we hope, become clear later in the chapter. ♦
Given a point xo 6 U c R", to evaluate a general 2-form on the ordered pair
(a, b) of vectors, we have
wXo(a, b) = Fi2(x0)<ixi A dx2(a, b) + Fi3(x0)<ixi A Jx3(a, b)
+ h F„_i„(x0)(ix„_i A dx„(a, b).
EXAMPLE 4 In R3, let co = 3xy dy A dz + (2y + z) dz A dx + (x - z) dx A
dy. Then
W(i.2,-3)(a, b) = 6 dy A dz(&, b) + dz A Jx(a, b) + 4 dx A <iy(a, b)
«3
fc3
+
#3
&3
/'I
+ 4
«2
&1
fe2
= 6(a2^3 - 03^2) + (^3^1 - aibi) + 4(aiZ?2 - «2^i)-
8.1 I An Introduction to Differential Forms 533
Finally, we generalize the notions of 1 -forms and 2-forms to provide a defi-
nition of a ft-form.
DEFINITION 1 .1 Let k be a positive integer. A basic (differential) A-form
on R" is an expression of the form
dxj{ A dxj2 A ■ ■ ■ A dxik ,
where 1 < ij < n for j = 1 , . . . , k. The basic &-forms are also functions that
require k vector arguments ai, aj, . . . , a* and are evaluated as
dx^i&i) dxh{2Lz) ■■■ dxh{ak)
dxix A ■ ■ ■ A dxik(n\ , . . . , a^) = det
dxi2{n\) dxi2(a2) ■ ■ ■ dxj2(ak)
dxit(a\) dxit(a2) ■■■ dxit(ak)
EXAMPLE 5 Let
ai = (1,2, -1,3,0), a2 = (5, 4, 3, 2, 1), and a3 = (0, 1, 3, -2, 0)
be three vectors in R5. Then we have
dx\ A dxT, A dx5(a\, a2, a^) = det
Using properties of determinants, we can show that
1 5 0
1 3 3
0 1 0
dxi, A ■ ■ • A dxj. A • • ■ A dxi, A • • ■ A dXik
(3)
= —dXjl A ■ • • A dXjt A • • • A dXjj A • • • A dXik
and
dxix A ■ • ■ A dX[ A ■ ■ • A dXjj A ... A dXjk = 0.
(4)
Formula (3) says that switching two terms (namely, dxi and dxt,) in the basic
&-form dxix A • • • A dxik causes a sign change, and formula (4) says that a basic
&-form containing two identical terms is zero. Formulas (3) and (4) generalize
formulas (1) and (2).
DEFINITION 1 .2 A general (differential) A-form on U C R" is an expres-
sion of the form
n
co = Fii ^jk(x)dxii A ■ • ■ A dxik,
i'i,...,i't=l
where each f^,,.*, is a real-valued function F,, ..,t: [7 — >• R. Given a point
xo £ {/, we evaluate to on an ordered £-tuple (ai , . . . , a*) of vectors as
n
coXo(au...,ak)= ^2 Fh-ikix<y)dxh A--- Adxik{ax, . . . ,ak).
h,...,ik—l
534 Chapter 8 | Vector Analysis in Higher Dimensions
Note that a 0-form is so named because, in order to be consistent with a
1-form or 2-form, it must take zero vector arguments!
In view of formulas (3) and (4), we write a general /c-form as
a> = ^2 Fh-k^xh A • • • A dxik.
\<i\ <---<ifr<n
(That is, the sum may be taken over strictly increasing indices i\, . , . , 4.) For
example, the 4-form
co = xi dx\ A dx?, A dx\ A dx$ + {xj, — x\)dx\ A dx-± A dx$ A dxi,
+ X\X^ dx$ A t/x3 A dx4 A Jxi
may be written in the "standard form" with increasing indices as
CO = (X2 — X\Xi)dx\ A <ix3 A dx4 A dxs + (xj — X3) dx\ A dx2 A dx3 A dx$.
Two &-forms may be added in the obvious way, and the product of a 0-form
/ and a &-form co is analogous to the product of a 0-form and a 1-form.
Exterior Product
The symbol A that we have been using does, in fact, denote a type of multiplication
called the exterior (or wedge) product. The exterior product can be extended to
general differential forms in the following manner:
DEFINITION 1.3 Lett/ c R" be open. Let / denote a 0-form on U . Let co =
dx, A • • • A dxik denote a &-form on U and 17 = J] Gjl..jldxj1 A
• • • A dxj, an /-form. Then we define
/ A co = fco = fFh...hdxh A ■ • ■ A dx,t,
CO A r\ = Fii...ikG j1...jldXjl A • • ■ A <£tjs A dXj, A ■ ■ • A fi?X/;.
Thus, the wedge product of a &-form and an /-form is a (& + /)-form.
EXAMPLE 6 Let
co = x\ dx\ A dx2 + (2x3 — X2)dx\ A JX3 + exi dxj A JX4
and
T] = X4 Jxj A JX3 A JX5 + X6 JX2 A JX4 A dX(,
be, respectively, a 2-form and a 3-form on R6. Then Definition 1.3 yields
co A rj = x\xa, dx\ A dx2 A Jxj A JX3 A dx$
+ (2X3 — X2)X4 dx\ A JX3 A dx\ A C/X3 A dx$
+ eX3X4 JX3 A dX4 A rfxi A rfx3 A dx$
+ X^X6 Jxj A dX2 A dX2 A dX4 A dxg
+ (2X3 — X2)X(, dx\ A JX3 A dX2 A C/X4 A dx(,
+ eA3X6 JX3 A dX4 A C/X2 A C/X4 A dx§.
8.1 | Exercises 535
Because of formula (4), most of the terms in this sum are zero. In fact,
co A ?7 = (2x3 — X2)x(> dx\ A dxi, A dxi A dx4 A dxc,
= (X2 — 2xt, )X6 dx\ A dX2 A dx?, A dx4 A dX(,,
using formula (3). ♦
From the various definitions and observations made so far, we can estab-
lish the following results, which are useful when computing with differential
forms:
PROPOSITION 1 .4 (Properties of the exterior product) Assume that all
the differential forms that follow are defined on U c R" :
1. Distributivity. If co\ and 002 are &-forms and r\ is an /-form, then
(&>i + 002) A Y] = co\ A rj + 0J2 A rj.
2. Anticommutativity. If co is a &-form and rj an /-form, then
CO AT] = (— \)H7] A CO.
3. Associativity. If co is a k-foxm, rj an /-form, and x a p-form, then
(a> A rj) A x = co A (rj A x).
4. Homogeneity. If co is a /c-form, an /-form, and / a 0-form, then
(fco) A rj = f(co A rj) = co A (frj).
8.1 Exercises
Determine the values of the following differential forms on the
ordered sets of vectors indicated in Exercises 1—7.
1. dx\ — 3dx2; a = (7, 3)
2. 2dx + 6dy-5dz; a = (1,-1,-2)
3. 3dxi a dx2; a = (4, -1), b = (2, 0)
4. Adx Ady -Idy A dz; a = (0, 1, -1), b = (1, 3, 2)
5. IdxAdyAdz; a = (1,0, 3), b = (2,-l,0), c =
(5,2, 1)
6. dx\ A dx2 + 2dx2 A t/x3 + 3^x3 A dx^, a = (1, 2,
3,4), b = (4, 3,2, 1)
7. 2dxi A d^3 A dx4 + dx2 A JX3 A dx$; a = (1, 0,
-1, 4, 2), b = (0, 0, 9, 1, -1), c = (5, 0, 0, 0, -2)
8. Let w be the 1-form on R3 defined by
co = x2y dx + y2z dy + z3x dz.
Find &)(3,_ii4)(a), where a = (a\,ai, a{).
9. Let co be the 2-form on R4 given by
co = X1X3 dx\ A dx?, — X2X4 dx2 A dx4.
Find w(2,_i,_3,i)(a, b).
10. Let co be the 2-form on R3 given by
co = cosxJxA dy — sinz dyA dz + (y2 + 3)dxA dz.
Find <W(o,-i,jr/2)(a> b), where a = (ai,a2,aj,) and
b = (bi,b2X).
11. Let co be as in Exercise 10. Find W(t y z)((2, 0, —1),
(1,7,5)).
1 2. Let co be the 3-form on R3 given by
co = (ex cos y + (y2 + 2)e2z) dx A dy A dz.
Find o>(o,o,o)(a, b, c), where a = (ai, a.2, 03), b =
(bub2, bi), and c = (ci, c2, c3).
13. Let a; be as in Exercise 12. Find co(x y ^((\,Q,Q),
(0, 2, 0), (0, 0, 3)).
In Exercises 14—19, determine co A rj.
14. On R3: co = 3 dx + 2dy — x dz; rj = x2dx —
cosydy + 1 dz.
1 5. On R3 : co = y dx — x dy; rj = z dx A dy + y dx A
dz + x dy A dz.
536
Chapter 8 | Vector Analysis in Higher Dimensions
16.
17.
21. Prove formula (4). (Hint: Use formula (3).)
22. Explain why a fc-form on R" with k > n must be iden-
tically zero.
23. Prove property 1 of Proposition 1.4.
18.
On R4: co = (x\ + X2)dx\ A dx% A dxi + (x-j — X4)
dx\ A dx2 A dxi\\ rj = x\ dx\ + 2x2 dx2 + 3x3 dx$.
24. Prove property 2 of Proposition 1 .4. (Hint: Use for-
mula (3).)
19.
On R5: co = x\ dxi A dx^ — X2X3 dx\ A dx$;
rj = eXiXidx\ A dx4 A dx^ — X\ COSX5 dx2 A dx^ A
dx4.
25. Prove property 3 of Proposition 1.4.
26. Prove property 4 of Proposition 1.4.
20.
Prove formula (3) by evaluating dxi{ A dxi2 A • • • A
dxik on k vectors ai, . . . , a* in R".
8.2 Manifolds and Integrals of k -forms
In this section, we investigate how to integrate &-forms over ^-dimensional objects
(i.e., curves, surfaces, and higher-dimensional analogues) in R".
Integrals over Curves and Surfaces
We begin by considering integrals of 1 -forms and 2-forms over parametrized
curves and surfaces.
DEFINITION 2.1 Let x: [a, b] R" be a C1 path in R". If co is a 1-form
defined on an open set U c R" that contains the image of x, then the integral
of co over x, denoted fx co, is
EXAMPLE 1 Let co = (x2 + y) dx + yz dy + (x + y - z) dz. We integrate co
over the path x: [0, 1] R3, x(t) = (2t + 3, 3r, 7 - t).
We have x'(t) = (2, 3, — 1) so that, using Definition 2.1, we find that
0=1 a>(2t+3,3t,7-o(2» 3, -l)dt
Jo
+ (It + 3 + 3t - (7 - t))dz(2, 3,-1)] rfr
♦
In general, if
co = F\ dx\ + F2 dx2 + h Fn dx,
8.2 | Manifolds and Integrals of k-forms 537
is a 1-form on R" and x: [a, b] — > R" is any path, then
«xW(x'(0) = F1(x(0)^1(x'(0) + F2(x(t))dx2(x'(t))
+ ■■■ + Fn(x(t))dxn(x'(t))
= FiixQMit) + F2(x(/))4(0 + ■ ■ ■ + F„(x(0K(f)
= (F1(x(0),F2(x(0),...,Fn(x(0))-x'(0.
From this we conclude the following:
PROPOSITION 2.2 If F denotes the vector field (FltF2,..., Fn) and
co = F\ dx\ + F2 dx2 -\ + Fn dx„
and if x: [a, b] — > R" is a C1 path, then
fco=f cox(n(x'(t))dt = f F(x(t))-x'(t)dt = f F-ds.
That is, integrating a 1-form over a path (or, indeed, over a simple, piecewise C
curve) is exactly the same as computing a vector line integral.
Now we see how to integrate 2-forms over parametrized surfaces in R3 .
DEFINITION 2.3 Let D be a bounded, connected region in R2 and let
X: D — » R3 be a smooth parametrized surface in R3. If co is a 2-form defined
on an open set in R3 that contains X(D), then we define fx on, the integral
of co over X, as
(Recall that T.v = 3X/3<f and T, = dX/dt.)
Let's work out the integral in Definition 2.3. We write co as
F\ dy Adz + F2 dz A dx + F3 dx A dy
and X(s, /) as (x(s, t), y(s, t), z(s, t)). Therefore,
1 1 [Fi(X(s,t))dy Adz(Ts,T,)+ F2(X(s,t))dz Adx(Ts,Tt)
J JD
+ F3(X(s, t)) dx A dy(Ts , Tr)] ds dt.
By definition of the basic 2-forms,
dy A Jz(TJ;T,) = det
dy(Ts)
dz(Ts)
dy(Tt)
dz(Tt)
= det
dy/ds
dz/ds
dy/dt
dz/dt
3(y,z)
d(s, t)
Chapter 8
Vector Analysis in Higher Dimensions
Similarly, we have
3(z, x) d(x, y)
dz A dx(Ts, Tf) = and dx A dy(Ts, T,) =
d(s,t)
Hence, if F = (F1; F2, F3), then
3(J, 0
W/.[
Fi(X(jr, 0)^7 r + F2(X(s, t))-
+ F3(X(s,t))
d(s, t)
d(x, y)
d(s, t)
9(5, 0
ds dt
-a
D V 3(M) 3(M) 9(M).
Recall from formula (7) in §7.1 that
3(y, z) 3(z, ■*) 9(*. y)
3(s,0' 3(5, 0 ' 3(s, 0
N(s, 0,
the normal to X(Z)) at the point X(s, f). Therefore, we have established the fol-
lowing Proposition (see also Definition 2.2 of Chapter 7):
PROPOSITION 2.4 If F denotes the vector field F = Fj i + F2 j + F3 k and
a) = F\ dy A dz + F2 dz /\ dx + F3 A dy
and if X: £) — » R3 is a smooth parametrized surface such that co (or F) is defined
on an open set containing X(D), then
L--II.
<WxCj,o(Ts, T,)dsdt
II.
-IL
F(X(s, 0)-N(s, 0^
Parametrized Manifolds
Next, we generalize the notions of parametrized curves and surfaces to
higher-dimensional objects in R". To set notation, let R^ have coordinates
(U\, U2 Wjfc).
DEFINITION 2.5 Let D be a region in RA that consists of an open, connected
set, possibly together with some or all of its boundary points. A parametrized
A-manifold in R" is a continuous map X: D — >• R" that is one-one except,
possibly, along 3D. We refer to the image M = X(D) as the underlying
manifold of X (or the manifold parametrized by X).
Such a & -manifold possesses k coordinate curves defined from X by
holding all the variables U \ , . . . , Ufa fixed except one; namely, the jth coor-
dinate curve is the curve parametrized by
uj 1 — > X(a\, . . . , dj-i, Uj, Uj+\, . . . , at),
8.2 | Manifolds and Integrals of k-forms 539
where the a, 's (i ^ j ) are fixed constants. If X is differentiable and x\ , X2
xn denote the component functions of X, then the tangent vector to the jth
coordinate curve, denoted TM , is
3X / dx\ 3x2 9x„
j =
' duj \duj ' duj ' ' duj
A parametrized k -manifold is said to be smooth at a point X(uo) if the
mapping X is of class C1 in a neighborhood of Uo and if the k tangent vectors
TMl , . . . , TUl are linearly independent at X(uo). (Recall that k vectors vi , . . . , v&
inR" are linearly independent if the equation civi + • • • + c^k = 0 holds if and
only if c\ = c2 = ■ ■ ■ = ck = 0.) A parametrized k -manifold is said to be smooth
if it is smooth at every point of X(uo) with uo in the interior of D.
Sometimes we will refer to the underlying manifold M = X(D) of a para-
metrized manifold X: D — > R" as a parametrized manifold; we do not expect any
confusion will result from this abuse of terminology.
EXAMPLE 2 Let D = [0, 1] x [1, 2] x [-1, 1] andX: D -+ R5 be given by
X(wi, U2, W3) = (Ml + U2, 3«2> W2W3, U2 — M3, 5^3).
We show that M = X(D) is a smooth parametrized 3-manifold in R5.
Note first that X is continuous (in fact, of class C°°) since its component
functions are polynomials. To see that X is one-one, consider the equation
X(u) = X(u); (1)
we show that u = u. Equation (1) is equivalent to a system of five equations:
U\ + U-2 = U\ + U2
3U2 = 3^2
W2«3 = M2M3
U2 — W3 = U2 — W3
5«3 = 5^3
The second equation implies u2 = U2, and the last equation implies u3 = S3.
Hence, the first equation becomes
U\+U2 = U\+U2 <S=> U\=U\.
Thus,
u = {u\, U2, M3) = (Hi, ui, U3) = u.
To check the smoothness of M, note that the tangent vectors to the three
coordinate curves are
TUI = — =(1,0, 0,0,0);
au\
9X
T„2 = — = (1,3,4 1.0);
ax
T„3 = - — = (0, 0, 2w2m3, -1, 5).
d«3
540 Chapter 8 | Vector Analysis in Higher Dimensions
Therefore, to have ciTi + C2T2 + C3T3 = 0, we must have
(ci + C2, 3c2, u\c2 + 2M2M3C3, C2 — C3, 5c3) = (0, 0, 0, 0, 0).
It readily follows that c\ = c2 = C3 = 0 is the only possibility for a solution.
Hence, T„, , T„2, T„3 are linearly independent at all u e D and so M is smooth at
all points. ♦
Figure 8.1 The planar robot arm
of Example 3 . Each rod is free to
pivot about the appropriate linkage
points.
Parametrized k -manifolds, although seemingly abstract mathematical notions
when k is larger than 3, are actually very useful for describing a variety of situa-
tions, one of which is illustrated in the next example.
EXAMPLE 3 A planar robot arm is constructed consisting of three linked rods
of lengths 1, 2, and 3. (See Figure 8.1.) The rod of length 3 is anchored at the
origin of R2 but free to rotate about the origin. The rod of length 2 is attached to
the free end of the rod of length 3, and the rod of length 1 is, in turn, attached to
the free end of the rod of length 2. We describe the set of positions that the arm
can take as a parametrized manifold.
Clearly, each state of the robot arm is determined by the coordinates (x\ , yi ),
(*2 , yi), and (x3 , y3) of the linkage points, which we may consider to form a vector
x = (jci, y\, x2, y2, *3, V3) in R6. However, not all vectors in R6 represent a state
of the robot arm. In particular, the point (x\ , yi) must lie on the circle of radius 3,
centered at the origin, the point (x2 , y2) must lie on the circle of radius 2, centered
at (jti, yi), and the point (x3, y3) must lie on the circle of radius 1, centered at
(x2, y2). Thus, for x = (jti, y\, x2, yi, yi) to represent a state of the robot arm,
we require
x\ + y\ = 9
(x2 - Xl)2 + (y2 - yi)2
(x3 - X2f + (V3 - V2)2
4.
1
(2)
and
We may parametrize each of the circles in the system (2) in a one-one fashion by
using three different angles 9\, 02, and ©3. Hence, we find
(jci, yi) = (3 cos 0i, 3 sin0i),
(x2, y2) = (x\ + 2 cos 02, yi + 2 sin02)
= (3 cos 0i + 2 cos 02, 3 sin 0! +2sin02), (3)
(*3» y3) = O2 + cos 03, y2 + sin03)
= (3 cos 0i+2 cos 02 + cos 03 , 3 sin 0i + 2 sin 02 + sin 03),
whereO < 0i, 02 , 03 < 27T . Therefore, the map X: [0, 27r) x [0, 27i) x [0, 27r) — >
R6 given by
X(0i, 02 , 03) = (xi,yi,x2, y2, x3, y3),
where (xi, y\, x2, y2, x3, y3) are given in terms of 0i, 02, and 03 by means of the
equations in (3), exhibits the set of states of the robot arm as a parametrized 3-
manifold in R6. We leave it to you to check that X defines a smooth parametrized
3 -manifold. ♦
8.2 | Manifolds and Integrals of k-forms 541
Just like a parametrized surface, a parametrized ^-manifold M = X(D) may
or may not have a boundary, denoted as usual, by dM. If M has a nonempty
boundary, then dM is contained in the image under X of the portion of the
boundary of the domain region D that is also part of D. Under suitable (and
mild) hypotheses, dM, if nonempty, is, in turn, a union of finitely many (k — 1)-
manifolds (without boundaries).
EXAMPLE 4 Let B c R3 denote the closed unit ball {u = (hj , w2, h3) | u\ +
u\ + u\< 1}, and define X: B -> R4 by
Then M = X(B) is a portion of a "generalized paraboloid" having equation w =
x2 + y2 + z2; we have M = {(x, y, z, w) £ R4 I w = x2 + y2 + z2, x2 + y2 +
z2 < 1}. In this case, dM = [(x, y, z, 1) | x2 + y2 + z2 = 1}. Note that dM is a
parametrized 2-manifold in R4, as we may see via the map
Y: [0, 7r] x [0, 2tt) — > R4, Y(i, t) = (sins cost, sins sin?, coss, 1). ♦
Integrals over Parametrized ^-manifolds —
Now, we see how to define the integral of a k-ioxm over a smooth parametrized
A- -manifold. Our definition generalizes those of Definitions 2.1 and 2.3.
DEFINITION 2.6 Let D be a bounded connected region in R* and X: D
R" a smooth parametrized ^-manifold. If is a &-form defined on an open
set in R" that contains M = X(D), then we define the integral of 00 over M
(denoted fx co) by
(Here f ■ ■ ■ f refers to the ^-dimensional integral over D.)
EXAMPLE 5 Let X: [0, 1] x [1, 2] x [-1, 1] -> R5 be the parametrized 3-
manifold defined by
(See Example 2.) Let co be the 3 -form defined on R5 as
co = X1X3 dx\ A dxj, A dxs + (X3X4 — 2x2x5) dx2 A dx4 A JX5.
We calculate fx co.
Recall from Example 2 that the tangent vectors to the three coordinate curves
are
X(«i, M2, W3) = (hi, H2, H3, u\ + u\ + h2).
X(Hi, H2, M3) = (Hi + H2, 3H2, M2H3, H2 — H3, 5H3).
T
T
(1,0, 0, 0, 0),
(1,3,h2, 1,0),
and
T,
(0, 0,2h2h3, -1,5).
542 Chapter 8 | Vector Analysis in Higher Dimensions
Then, from Definition 2.6,
L
= + u2)u2u\dxi Adx3 Adxs(Tul, TU2, TM3)
J-\ Ji Jo
+ (u2u\(u2 — M3) — 30^2^3)^2 A ^4 A ^X5(TM1, T„2, T„3)} du\ du2 duT,
f-i Ii lo
(ui + u2)u2u3
1 1
0 W3 2m2"3
0 0
0
3
0
+ (M2M3 — W2M3 — 3OM2M3)
0
1
-1
0
0
5
du\ du,2 duT,
/I I 5(ui + U2)U2u\ du\ du2 du-i,
1 J\ Jo
37
6 ■
EXAMPLE 6 If <w is a 3-form on R3 , then co may be written as
co = F(x, y, z)dx A dy A dz.
(Why?) If D* is a bounded region in R3 and X: D* -> R3 is a smooth parametrized
3-manifold, then Definition 2.6 tells us that
j 0} = J J j COx(uuui:u3)(J'ui,Tu2>r^u3)dUldu2dU3
= III F(X(u))dx A dy A dz(TUl, TM2, T„3)<iMi du2 dui
dx/dui dx/du2 dx/dui,
dy/dui dy/du2 dy/dui,
dz/dui dz/du2 3z/3w3
du\ du2 du-i
d(x, y, z)
F(x(u), y(u), z(u))— du\ du2 du3
D* d{uUU2,Uyj
III
± j j j F(x, y, z)dx dy dz,
from the change of variables theorem for triple integrals (Theorem 5.5 of Chapter
5), where D = X(D*). ♦
Orientation of a Parametrized Zj-manifold
We have seen that vector line integrals and vector surface integrals may be defined,
respectively, over oriented curves and surfaces in a manner effectively independent
of the parametrization used. We now see how it is possible to define the integral
8.2 | Manifolds and Integrals of k-forms 543
C
Figure 8.2 An orientation of the
curve C shown is a choice of
continuously varying unit tangent
vector T along C.
N N
Figure 8.3 An orientation of
the surface S is a choice of
continuously varying unit normal
vector N along S.
of a £-form over a parametrized & -manifold X: D —> R" so that it depends largely
on the underlying manifold M = X(D), rather than on the particular map X. To
do this, we must consider how reparametrization of M affects the integral, and
we must define what we mean by an orientation of M.
First, we consider the notion of orientation. We have previously seen how
parametrized curves and surfaces can be oriented by using some fairly natural
geometric ideas. A smooth parametrized curve implicitly received an orientation
from the parameter; typically, we orient a curve by indicating the direction in
which the parameter variable increases. We may also think of an orientation of
a curve as a choice of a unit tangent vector T at each point of the curve, made
so that T varies continuously as we move along the curve. (See Figure 8.2.) An
orientation of a smooth parametrized surface in R3, when it exists, is a choice
of a continuously varying unit normal vector N at each point of the surface. (See
Figure 8.3.)
To define notions of orientation and orientability for a parametrized k-
manifold when k > 2, we will need to work more formally.
First, we need to introduce two related ideas from the linear algebra of R" .
Thus, suppose Vi, \i, . . . , \k are vectors in R". By a linear combination of
Vi , . . . , Vfc, we mean any vector v e R" that can be written as
v = civi +c2v2H \-ckvk
for suitable choices of the scalars c\ , . . . ,ck. The set of all possible linear com-
binations of vi V£, called the (linear) span of vi , . . . , \k, will be denoted
Span{vi , . . . ,\k}. That is,
Span{v! ,
{civi H \-ck\k \ci,...,cke R}.
DEFINITION 2.7 Let M = X(D), where X: D C R* R", be a smooth
parametrized k -manifold. An orientation of M is a choice of a smooth,
nonzero &-form Q defined on M. If such a &-form £1 exists, M is said to be
orientable and oriented once a choice of such a A'-form is made.
Although we cannot readily visualize an orientation £2 of a parametrized k-
manifold when k is large, we can nonetheless see how the tangent vectors to the
coordinate curves relate to it.
DEFINITION 2.8 Let M = X(D) be a smooth parametrized k -manifold
oriented by the A-form Q. The tangent vectors TUl , . . . , T„4 to the coordinate
curves of M are said to be compatible with Q if
^x(u)(Tul
T„J > 0.
We also say that the parametrization X is compatible with the orientation Q
if the corresponding tangent vectors TUl
, T„, are.
Note that if TMl TUk are incompatible with the orientation Q, then they
are compatible with the opposite orientation — £1. Alternatively, we may change
the parametrization X of M by reordering the variables u\, . . . , uk to, say, 112, ui,
«3,
sothatT„2,TUl,
TUk are compatible with Q..
544 Chapter 8 | Vector Analysis in Higher Dimensions
Definition 2.7 is consistent with the earlier definitions of orientations of
curves and surfaces, as we now discuss. Suppose first that x: / -> R" is a smooth
parametrized curve in R" (where / is an interval in R) and T is a continuously
varying choice of unit tangent vector along C = x(7). Then we may define an
orientation 1-form £2 on C by
^x(r)(a) = T • a.
Conversely, given an orientation 1-form Q, we may define a continuously varying
unit tangent vector from it by taking T to be the unique unit vector parallel to
x'(0 such that, for any nonzero vector a parallel to x'(?),
T • a has the same sign as £2x(f)(a).
That T is uniquely determined follows because T must equal ±xr(t)/\\x'(t)\\, so
knowing a and the value of £2x(f)(a) determines the choice of sign for T.
Similarly, suppose 5 = X(D) is a smooth parametrized surface in R3 (i.e., a
smooth parametrized 2-manifold). If we can orient S by a continuously varying
unit normal N, then we may define an orientation 2-form Q on 5 by
fiX(«1,«2)(a,b) = det[ Nab],
where [ N a b ] is the 3x3 matrix whose columns are, in order, the vectors
N, a, b. Conversely, given an orientation 2-form Q on 5, we may define a con-
tinuously varying unit normal N from it by taking N to be the unique unit vector
perpendicular to T„, and T„2 (and hence to every vector in Span{T„, , T„2}) such
that, for any pair a, b of linearly independent vectors in SpanjT,,, , T„2},
det [ N a b ] has the same sign as £2X(i(1,„2)(a, b).
To see that N is uniquely determined, note that, given linearly independent vectors
a, b in Span{TI(1 , T„2}, the only possibilities for N are
T x T
I U 1 H2
|T„. x Tu
Hence, we choose the sign for the normal vector N so that det [Nab] has
the same sign as f2x(Ml,M2)(a, b).
EXAMPLE 7 Consider the generalized paraboloid M = {(x, y, z, w) e R4 |
w = x2 + y2 + z2}, which we may exhibit as a smooth parametrized 3-manifold
via
X: R3 — > R4, X(«i, U2, w3) = {u\, u%, m3, u\ + u\ + u2).
We show how to orient M.
Note that the equation x2 + y2 + z2 — w = 0 shows that M is the level set at
height 0 of the function F(x, y, z, w) = x2 + y2 + z2 — w. Hence, the gradient
V F = (2x ,2y,2z, — 1 ) is a vector normal to M . If we employ the parametrization
X and normalize the (parametrized) gradient, we see that
(2wi,2k2,2w3, -1)
N(Mi, U2, M3) =
' 4u2 + Au\ + Au\ + 1
is a continuously varying unit normal. Moreover, the 3-form Q, defined on M as
^x(u)(ai, a2, a3) = det [N ai a2 a3]
8.2 | Manifolds and Integrals of k-forms 545
gives an orientation for M. Note that
= det
I T„
T„ T„ 1
1 Aj,^- _l_ ^.ij2
T 1
^Au\
+ Au\ + Au\
+ 1
^Au\
+ Au\ +
+ 1
-1
^Au\
+ Au\ +
+ 1
+ 4u22
+ 4M2+ 1.
1 0 0
0 0
2wi 2u2 2«3
Since this last expression is strictly positive, we see that TM1, T„2, TU3 are com-
patible with £2. ♦
EXAMPLE 8 We may generalize Example 7 as follows:
Suppose that M c R" is the graph of a function /: U C R"-1 R"; that
is, suppose M is defined by the equation jc„ = /(xi , . . . , x„_i). Then M may be
parametrized as an (n — 1 )-manifold via
Kn-l))-
X:U c R" 1 -> R\ X(Ml, ...,uB_i) = (mi,..., w„_i, /(m,
Since Af is also the level set at height 0 of the function
F(x\, . . . , xn) = f(X\, . . . , x„_i) — xn,
a vector normal to Mis provided by the gradient VF = (fXl, . . . , fx„_lt — 1). Ifwe
normalize VF and use the parametrization X, we see that we have a continuously
varying unit normal
N(m
, Un-\)
{fill I • • • ! fll„-l ! 1)
from which we may define our orientation (n — l)-form £2 for M by
fix(u)(ai,...,a„_i) = det[N ai ■•■ a„_i].
Now suppose that M is a smooth parametrized £ -manifold in R" with non-
empty boundary dM. If M is oriented by the &-form £2, then there is a way to
derive from it an orientation for dM, which we describe in Definition 2.9. To
set notation, let X: D C R* — >• R" denote the parametrization of M and suppose
Y: E c RA'~' — R" gives a parametrization of a connected piece of 3M as a
smooth (k — l)-manifold. Since dM is part of M, if s = (si , . . . , S£_i) e E, then
there is some u = (u\, . . . , Uk) e D such that Y(s) = X(u).
546 Chapter 8 | Vector Analysis in Higher Dimensions
DEFINITION 2.9 Let M be a smooth parametrized /: -manifold in R" with
boundary dM. Suppose M is oriented by the &-form Q. Then the connected
pieces of dM are said to be oriented consistently with M, or that dM has
its orientation induced from that of M , if the orientation (k — l)-form Q3M
is determined from Q as follows. Let V be the unique, outward-pointing unit
vector in R", defined and varying continuously along dM, that is tangent to
M and normal to dM. (See Figure 8.4.) Then £2aM is defined as
n£«(«i. . . . , ak-i) = C2X(u)(V, alt . . . , aw),
where the map X: D c R* ^ R" parametrizes M, the map Y: E c Rk~l ->
R" parametrizes a connected piece of dM, and Y(s) = X(u).
Note that, in particular, the vector V in Definition 2.9 must be such that
• V e Span{T„, , . . . , T„J (i.e., V is tangent to M);
■ V- TSj = 0 for i = 1, . . . , k - 1 (i.e., V is normal to dM);
' V points away from M.
These conditions are often not difficult to achieve in practice. Definition 2.9 will
be very important when we consider a generalization of Stokes's theorem in the
next section.
EXAMPLE 9 Consider the surface 5 in R3 consisting of the portion of the
cylinder x2 + y2 = 4 with 2 < z < 5. Note that the boundary of S consists of the
two circles {(x, y, z) \ x2 + y2 = 4, z = 2}and{(x, y, z) \ x2 + y2 = 4, z = 5}.
We investigate how to orient 35 consistently with an orientation of S.
The cylinder may be parametrized as a 2 -manifold in R3 by
X: [0, 2tt) x [2, 5] — > R3, X(wi, ui) = (2cosmi, 2sinwi, uj).
Then the tangent vectors to the coordinate curves are
TMl = (— 2sinM!, 2coswi, 0)
and
TU2 = (0, 0, 1).
Since S is a portion ofthe level set at height 4 of the function F(x, y, z) = x2 + y2,
a unit normal N to S is given by
(2x, 2y, 0) = tx_ y \
l|VF|| J4x2 + 4y2 V2'2' )'
In terms of the parametrization X, the normal N is also given by
N = (cos Mi, sin Mi, 0).
Then we may define an orientation 2-form on S by
£2x(jn,«2)(ai, a2) = det[N ai a2].
Figure 8.4 The outward-pointing
unit vector V of Definition 2.9.
8.2 | Manifolds and Integrals of k-forms 547
Hence,
Figure 8.5 Orienting the
boundary of the surface S of
Example 9. Note the
outward-pointing tangent vectors
Vtop and Vbottom-
(THl, TH2) = det
COSM]
0
—2 siriM]
2 cos u\
0
2 > 0.
Thus, TH1, TH2 are compatible with £2.
We may parametrize 35 by using two mappings:
and
Bottom circle: Yi : [0, 2n) -> R3
Top circle: Y2: [0, 2tt) -> R3
Yi(s) = (2 cos 5, 2 sins, 2)
Y2(s) = (2 coss, 2 sins, 5)
To use Definition 2.9 to orient 35, we must identify outward-pointing vectors tan-
gent to 5 and normal to 35. From Figure 8.5, we see that along the top circle V =
Vtop = (0, 0, 1) works, while along the bottom circle, V = Vbottom = (0,0,-1)
suffices. Hence, Definition 2.9 tells us that, along the bottom circle,
nYi(s)(a) = ^X(i,2)(Vbottom, a) = det [ N Vbottom 3 ] ,
while along the top circle,
^)(a) = Qx(,,5)(Vtop,a) = det[N Vtop a].
For both maps Yi and Y2, we have that the coordinate tangent vector is Ts =
(—2 sins, 2 cos s, 0). Thus, along the bottom circle,
Q
dS
YiCs)
(Ts) = det
coss
sins
0
0
0
-1
—2 sins
2 cos s
0
= 2,
so Ts is compatible with the orientation 1-form £ldS. However, along the top
circle,
~ coss 0 —2 sins
^(s)(Tv) = det sins 0 2coss
0 1 0
so Ts is incompatible with £2 . Therefore, we must orient the top circle clockwise
around the z-axis and the bottom circle counterclockwise. ♦
The following example is the three-dimensional analogue of Example 9:
EXAMPLE 10 Consider the subset M c R4givenbyM = {(x, y, z, w) \ x2 +
y2 + z2 = 4, 2 < w < 5}. This set M is a portion of the cylinder over a sphere
of radius 2. Note that the boundary of M consists of the two spheres 5bottom =
{(*, y, z, 2)\x2 + y2 + z2 = 4} and5top = {(*, y, z, 5) | x2 + y2 + z2 = 4}. We
investigate M and dM as parametrized manifolds, orient M, and study the induced
orientation on dM.
First, we note that M may be parametrized as a 3 -manifold in R4 by
X: [0, it] x [0, 2tt) x [2, 5] -> R4,
X(«i, «2, K3) = (2 sin mi cos U2, 2sinwi sinM2, 2cosmi, M3).
(This is the usual parametrization of a sphere using spherical coordinates (p = u\,
9 = 112, with an additional parameter M3 for the "vertical" w-axis.) The tangent
548 Chapter 8 | Vector Analysis in Higher Dimensions
vectors to the coordinate curves are given by
TUl = (2cos«i cos U2, 2cosmi sinM2, — 2sinwi, 0),
T„2 = (—2 sinw! sinM2, 2 sinwi cos 112, 0, 0),
and
TH3 = (0, 0,0, 1).
Note that this parametrization fails to be smooth when u\ is 0 or jt, since then
T„2 = 0 at those values for u \ . You can check that the parametrization is smooth
at all other values of X(u) (i.e., for u in (0, jt) x [0, 2jt) x [2, 5]).
Because M is a portion of the level set at height 4 of the function F(x, y,
z, w) = x2 + y2 + z2, a unit normal N to M is given by
VF (2x,2y,2z, 0) (x y z
l|VF|| J4x2 + 4y2 + 4z2 V2' 2' 2"
In terms of the parametrization X, the normal N is also given by
N = (sinwi cosw2, sinwi sinw2, cosmi, 0).
We define an orientation 3 -form Q for M by
^x(u)(ai,a2,a3) = det[N ai a2 a3].
Then
^x(u)(TM1, T„2, T„3) = det
sin u\ cos U2 2 cos Mi cos i<2 — 2 sin Mi sin M2 0
sinwisinw2 2cosMisinM2 2sinwicos«2 0
coswi —2 sin Mi 0 0
0 0 0 1
= 4 sin u i > 0
for 0 < wi < jt (which is where the parametrization X is smooth). Hence, THI,
T„2, T„3 are compatible with Q..
We parametrize the two pieces of dM with two mappings:
"Bottom" sphere Sbottom:
Yi:[0,tt] x [0,2tt)^ R4,
Yi(^i, 52) = (2 sin^i cos^2> 2sinsi sin52, 2cos5i, 2),
and
"Top" sphere Stop:
Y2:[0, jt] x [0,2;r)^ R4,
Y2(*i, S2) = (2 sinsi COSS2, 2sinsi sin 52, 2cossi, 5).
Note that both parametrizations Yi and Y2 give the same tangent vectors to the
corresponding coordinate curves, namely,
Tj, = (2 cos si cos 52, 2 cos si sin 52, —2 sinsi, 0)
and
TS2 = (— 2sin5! sin52, 2sin5! cos 52, 0, 0),
and by considering these tangent vectors, we see that the parametrizations are
smooth whenever 51 ^ 0, jt.
8.2 | Manifolds and Integrals of k-forms 549
To give dM the orientation induced from that of M, we identify outward-
pointing unit vectors tangent to M and normal to dM. Thus, we need V such that
• Ve Span{TM1,TM2,TM3} <=> V-N = 0;
• V-T.S1 =V-T.S2 = 0;
• V points away from M.
It's not difficult to see that we must take V = Vtop = (0, 0, 0, 1) along 5top and
V = Vbottom = (0, 0, 0, —1) along Sbottom- Therefore, Definition 2.9 tells us that
along Sbottom,
fi^s)(ai, a2) = ^x(s,2)(Vbottom, ai, a2).
In particular,
^s)(T.s.1,TJ2) = det[N Vbottom T„ TJ2]
= det
sin 5i cos 52 0 2 cos cos ^2 — 2 sin ji sin 52
sin^isin^ 0 2cossisini2 2sinsicoss2
cos^i 0 — 2sin^i 0
0-10 0
= 4 sin^i > 0
for 0 < s\ < 7T (i.e., where the parametrization Yj is smooth). Thus, Y\ is com-
patible with QdM . Along Stop, however, we have
^S)(T,1,TJ2) = det[N Vbottom Tn TJ2]
det
sin s\ cos 52 0 2 cos 5j cos 52 — 2 sin 5i sin 52
sin 5i sin 52 0 2cos5ishi52 2sin5iCOS52
cos 5i 0 —2 sin 5i 0
0 10 0
= — 4sin5i < 0
for 0 < 5i < it, so Y2 is incompatible with Q3M . We must take care with this
distinction when we consider the general version of Stokes's theorem. ♦
Next, we examine how the integral of a &-form co can vary when taken
over two different parametrizations X: D\ —> R" and Y: D2 —> R" for the same
^-manifold M = X(D0 = Y(D2).
DEFINITION 2.10 Let X: Di c R1 -> R" and Y: D2cR'^ R" be
parametrized k -manifolds. We say that Y is a reparametrization of X if there
is a one-one and onto function H: D2 —> D\ with inverse H 1 D\ — > D2 such
that Y(s) = X(H(s)), that is, such that Y = X o H. If X and Y are smooth and
H and H~ 1 are both class C 1 , then we say that Y is a smooth reparametrization
ofX.
Since H is one-one, it can be shown that the Jacobian det DH cannot change
sign from positive to negative (or vice versa). Thus, we say that both H and Y are
orientation-preserving if the Jacobian det DH is positive, orientation-reversing
if det DH is negative.
550 Chapter 8 | Vector Analysis in Higher Dimensions
The following result is a generalization of Theorem 2.5 of Chapter 7 to the
case of k -manifolds.
THEOREM 2.1 1 Let X: D,cR'^ R" be a smooth parametrized k -manifold
and co a k-form defined on X(£>i). If Y: Di c —> R" is any smooth
reparametrization of X, then either
if Y is orientation-preserving, or
if Y is orientation-reversing.
In view of Theorem 2.1 1, we can define what we mean by fM co, where M
is a subset of R" that can be parametrized as an oriented ^-manifold and is a
/c-form defined on M. We simply let
where X: D C R* -> R" is any smooth parametrization of M that is compatible
with the orientation chosen.
EXAMPLE 1 1 We evaluate fc co, where C is the (oriented) line segment in R3
from (0, — 1 , —2) to (1 , 2, 3) and co = z dx + x dy + y dz.
Using Theorem 2.1 1, we may parametrize C in any way that preserves the
orientation. Thus,
x: [0, 1] -> R3, x(0 = (1 - /)(0, -1, -2) + f(l, 2, 3) = (/, 3t - 1, 5t - 2)
is one way to make such a parametrization. Then x'(t) = (1, 3, 5) and hence,
from Definition 2.1, we have
co= co= CL>x(t)(x'(t))dt
JC Jx JO
= f {(5t - 2) • 1 + t • 3 + (3t - 1) • 5} dt
Jo
= f (23t-T)dt
Jo
23
It
Note that if we parametrize C in the opposite direction by using, for example,
the map
y: [0, 1] -> R3, y(f) = f(0, -1, -2) + (1 - 0(1, 2, 3) = (1 - f, 2 - 3r, 3 - 50,
8.2 | Exercises 551
then we would have
. i
f m= f a>y(t){y'(t))dt
Jy JO
= f {(3 - 5t)(- 1) + (1 - 0(~3) + (2 - 3f)(-5)} dt
Jo
= jf (23t - I6)dt = (^-t2 - I6t
In light of Theorem 2.11, this result could have been anticipated from our pre-
ceding calculation of fx co. ♦
Note on k -manifolds
The central geometric object of study in this section, namely, a parametrized
£ -manifold is actually a rather special case of a more general notion of a k-
manifold. In general, a A-manifold in R" is a connected subset M c R" such that,
for every point x e M, there is an open set U c R* and a continuous, one-one
map X: U — > R" with x e X(t/) c M. (A ^-manifold with nonempty boundary
requires a somewhat modified definition.) That is, M is a (general) k -manifold
if it is locally a parametrized & -manifold near each point. It is possible to extend
notions of orientation and integration of £-forms to this more general setting,
although it requires some finesse to do so. For the types of examples we are
encountering, however, our more restrictive definitions suffice.
8.2 Exercises
1 . Check that the parametrized 3 -manifold in Example 3 is in
fact a smooth parametrized 3 -manifold.
2. A planar robot arm is constructed by using two rods as
shown in Figure 8.6. Suppose that each of the two rods may
telescope, that is, that their respective lengths l\ and h may
vary between 1 and 3 units. Show that the set of states of
this robot arm may be described by a smooth parametrized
4-manifold in R4.
Figure 8.6 Figure for Exercise 2.
3. A planar robot arm is constructed by using a rod of length
3 anchored at the origin and two telescoping rods whose
respective lengths h and /3 may vary between 1 and 2 units
as shown in Figure 8.7. Show that the set of states of this
robot arm may be described by a smooth parametrized 5-
manifold in R6. (See Exercise 2.)
Figure 8.7 Figure for Exercise 3.
4. A robot arm is constructed in R3 by anchoring a rod
of length 2 to the origin (using a ball joint so that
the rod may swivel freely) and attaching to the free
end of the rod another rod of length 1 (which may
also swivel freely; see Figure 8.8). Show that the set of
states of this robot arm may be described by a smooth
parametrized 4-manifold in R6.
5. Suppose vi , . . . , V£ are vectors in R" . If x e R" is orthog-
onal to V; for i = 1 , . . . , k, show that x is also orthogonal
to any vector in Spanfvi , . . . , v*}.
Chapter 8 | Vector Analysis in Higher Dimensions
- (x2, y2, z2)
xi,yi, zi)
Figure 8.8 Figure for Exercise 4.
6. Let a, and c be positive constants and x: [0, it] — > R3
the smooth path given by x(f ) = (a cost, b sin /, ct). If
co = b dx — a dy + xy dz, calculate J co.
7. Evaluate fc co, where C is the unit circle x2 + y2 = 1,
oriented counterclockwise, and co = y dx — x dy.
8. Compute fc co, where C is the line segment in R"
from (0, 0, ... , 0) to (3, 3, ... , 3) and co = x\ dx\ +
x\ dx2 + • • • + x" dxn .
9. Evaluate the integral f„ co, where X is the parametrized
helicoid
X(s, f) = (s cosf, s sinr, t), 0 < s < 1, 0 < t < An
and
co = z dx A dy + 3 dz A dx — x dy A dz.
10. Consider the helicoid parametrized as
X(mi , ui) = (u\ cos3m2,"i sin3M2, 5M2),
0 < ui < 5, 0 < m2 < lit.
Let S denote the underlying surface of the helicoid and
let Q, be the orientation 2-form defined in terms of X
X(Ml,«2
)(a, b) = det
— 5sin3«2 a\ b\
5 cos 3;<2 a-2 b2
— 3u\ fl3 £>3
(a) Explain why the parametrization X is incompatible
with Q.
(b) Modify the parametrization X to one having the
same underlying surface 5 but that is compatible
with n.
(c) Alternatively, modify the orientation 2-form Q to £2'
so that the original parametrization X is compatible
with £2'.
(d) Calculate fs co, where co = z dx A dy — (x1 +
y2)dy A dz and S is oriented using £2.
11. Let M be the subset of R3 given by {{x, y, z) \ x2 +
y2 — 6 < z < 4 — x2 — y2}. Then M may be
parametrized as a 3 -manifold via
X: D— > R3; X(i<i , u2, 113) = (u\ cos u2,u \ sini<2, M3),
where
D = {(uu u2, m3) e R3 ] 0 < in < VI, 0 < u2 < lit,
u\ - 6 < m3 < 4 - u\}.
(The parameters u\, u2, and M3 correspond, respec-
tively, to the cylindrical coordinates r, 9, and z. Hence,
it is straightforward to obtain the aforementioned
parametrization. )
(a) Orient M by using the 3-form Q., where
^x(u)(a, b, c) = det [a b cl.
Show that the parametrization, when smooth, is
compatible with this orientation.
(b) Identify BM and parametrize it as a union of two
2-manifolds (i.e., as a piecewise smooth surface).
( c) Describe the outward-pointing unit vector V, vary-
ing continuously along each smooth piece of 3M,
that is normal to BM. Give formulas for it in terms
of the parametrizations used in part (b).
1 2. Calculate fs co, where S is the portion of the paraboloid
z = x2 + y2 with 0 < z < 4, oriented by upward-
pointing normal vector (— 2x, —2y, 1), and co =
ezdx A dy + y dz A dx + x dy A dz.
13. Calculate fs co, where S is the portion of the cylinder
4 with — 1 < y < 3, oriented by outward
xz + zz
normal vector (x , 0, z),and&> = z dx A dy + ey dz A
dx + x dy A dz.
1 4. Consider the parametrized 2-manifold
X: [1, 3] x [0, 2tt) —> R4, X(s,t)
= (-y/s cos t, V4 — s sin t , -v/s sin t , -J A — s cos t).
Find
{x\ + X4) dx\ A dx->, — (2x\ + 2x|) dx2 A dxs,.
15. Consider the parametrized 3-manifold
X: [0, 1] x [0, 1] x [0, 1] R4,
X(«l, U2, M3) = (Ml, M2, M3, (2u\ — M3) ).
Find
L
X2 dx2 A JX3 A dx4 + 2xiX3 Jxi A dx2 A rfx3.
8.3 I The Generalized Stokes's Theorem 553
8.3 The Generalized Stokes's Theorem
We conclude with a discussion of a generalization of Stokes's theorem that relates
the integral of a £-form over a k -manifold to the integral of a (k — l)-form over
the boundary of the manifold. Before we may state the result, however, we need
to introduce the notion of the exterior derivative of a &-form.
The Exterior Derivative —
The exterior derivative is an operator, denoted d, that takes differential &-forms
to (k + l)-forms and is defined as follows:
DEFINITION 3.1 The exterior derivative df of a 0-form / on U C R" is
the 1-form
At df A j.^ A V A
df = - — dx\ + - — dx2-\ h - — dxn.
ax\ 0x2 oxn
For k > 0, the exterior derivative of a &-form
co = Fk_ikdxh A ■ • • A dxh
is the (k + l)-form
dco = J^JjdFju,jk) A dx{x A ■ • • A dxit,
where dFiujt is computed as the exterior derivative of a 0-form.
EXAMPLE 1 If
f(xl,x2,xi,x4,x5,x6) = X\X2Xi> + X4X5X6,
then
df = x2xt, dx\ + dx2 + X\x2 dxj, + x^Xf, dx4 + X4X6 dx$ + X4X5 cfxg. ♦
EXAMPLE 2 If &) is the 1-form
00 = x\x2 dx\ + X2X3 + (2xi — x2) dxj,,
then
Jo) = d(x\x2) A Jxi + d(x2Xi) A Jx2 + d(2xi — x2) A <ix3
= (x2 dx\ + xi A dx\ + (X3 J%2 + X2 dx?,) A Jx2 + (2dxi — dx2) A dx$ .
Using the distributivity property in Proposition 1 .4 and the facts that dxj A dx\ = 0
and dxj A rfx;- = — dxj A dx,-, we have
da> = x\ dx2 A dx\ + X2 dx3 A dx2 + 2 Jxi A dx^ — dx2 A JX3
= — x\ dx\ A dx2 + 2 Jxi A JX3 — (X2 + 1) dx2 A dxj,. ^
Stokes's Theorem for k -forms
We now can state a generalization of Stokes's theorem to smooth parametrized
k -manifolds in R" .
THEOREM 3.2 (Generalized Stokes's theorem) Let D c Rk be a closed
bounded, connected region, and let M = X(D) be an oriented, parametrized k-
manifold in R". If dM ^ 0, let 3M be given the orientation induced from that
554 Chapter 8 | Vector Analysis in Higher Dimensions
of M. Let a) denote a (k — l)-form denned on an open set in R" that contains M.
Then
/ doo = I i
JM JdM
If 9M = 0, then we take faM co to be 0 in the preceding equation.
We make no attempt to prove Theorem 3.2. 1 Instead, we content ourselves
for the moment by checking its correctness in a particular instance.
EXAMPLE 3 We verify the generalized Stokes's theorem (Theorem 3.2) for the
2-form co = zw dx A dy, where M is the 3-manifold M = {(x, y, z, 10) e R4
w = x2 + y2 + z2, x2 + y2 + z2 < 1} oriented by the 3-form £2 corresponding
to the unit normal
N =
(2x,2y,2z, -1)
y^x2 + 4y2 + 4z2 + 1
The manifold M is a portion of the 3-manifold given in Example 7 of §8.2
and may be parametrized as
X:B
R
X(mj, U2, M3) = (U\, U2, «3, u\ + U2 + uf),
where B = {(u\, u%, M3) | u\ + u\ + u\ < 1}. Using this parametrization, we
have
TM1 = (1,0,0, 2u{)
T„2 = (0, 1,0, 2u2)
T„3 =(0,0, 1,2k3)
(2m 1, 2«2, 2«3, —1)
N =
4w2 + + + 1
so the orientation 3-form Q is given by
^x(u)(ai, a2, a3) = det [N ai a2 a3].
Example 7 of §8.2 shows that the parametrization X is compatible with this
orientation. Hence, we may use this parametrization without any adjustments
when we calculate fM dco.
The boundary of M is dM = {(x, y, z, w) \ x2 + y2 + z2 = w = 1} and may
be parametrized as
Y: [0, tt] x [0, 2n)
Then
and
T —
-1 Si —
R4, Y(ji, s2) = (sin^! C0S52, sins\ sms2, cos^i, 1).
(cos^i cos^2. cos ifi sin.?2, — sinsi, 0)
(— sinii sin^2, sin^i cos^, 0, 0).
For a full and rigorous discussion of differential forms and the generalized Stokes's theorem, see J. R.
Munkres, Analysis on Manifolds (Addison- Wesley, 1991), Chapters 6 and 7.
8.3 I The Generalized Stokes's Theorem
555
An outward-pointing unit vector V = (vi, v2, v-$, V4) tangent to M and normal to
dM must satisfy
• V-N = 0 along dM;
• V-T.S1 =V-T.S2 =0.
Along dM, we have
N = — =(2sinsi coss2, 2sinsi sins2, 2cossi, —1).
v 5
Thus, V must satisfy the system of equations
(2sinsi coss2)i>i + (2 sin s\ sin 52)^2 + (2cossi)i>3 — V4 = 0
(cossi COSS2M + (cos si sins2)u2 — (sinsi)^ = 0
— (sinsi sins2)i>i + (sinsi coss2)i>2 = 0
After some manipulation, one finds that the unit vector that satisfies these equa-
tions and also points away from M is
V = — =(sinsi cos S2, sinsi sins2, cossi, 2).
V 5
Then the induced orientation 2-form Q3M for dM is given by
^)(ai,a2) = nX(„)(V)ai,a2))
where X(u) = Y(s). In particular, we have
^)(T.sl,TS2) = det[N V T„ Tj
= det
r^sinsicoss2 ^sinsicoss2 COSS1COSS2 — sinsisins2
■^sinsisins2 4| sins 1 sin s2 cossisins2 sinsiCOSs2
v/5
COS Si
_ J_
COS Si
V5
-sin si
0
0
0
= sin si > 0
for 0 < si < Tt. Hence, the parametrization Y of dM, when smooth, is compatible
with the induced orientation, so we may use this parametrization to calculate
a m
CO.
Now we are ready to integrate. We first compute fM da>. Since co = zw dx A
dy, we have
dm = d{zw) A dx A dy = (z dw + w dz) A dx Ady
= z dw A dx A dy + w dz A dx A dy.
Thus,
/ dco= dwx(u)(TUl, T„2, Tm)dui du2du3
Jm J J Jb
-III
M3 dw A dx A dy(Tm , TH2, T„3)
+ {u\ + u\ + u\)dz A dx A dy(TUl, T„2, T„3)} du\ du2 duj,
Chapter 8 | Vector Analysis in Higher Dimensions
Iff,
"3
lit
1
0
0
1
=2l(2
+ (Mj + u\ + u\)
0 0 1
1 0 0
0 1 0
= 1
du\ du2 duj,
J J J (u\ + u\ + 3u\)du\dii2 duT,.
Since B is a solid unit ball, the easiest way to evaluate this iterated integral is to
use spherical coordinates p, (p, and 9. Hence,
n n7.7t p7t n\
I dco= I I I (p2 + 2p2 cos2 cp) p2 sirup dpdcpdO
Jm Jo Jo Jo
= / / / p4 (sirup + 2 cos2 <p simp) dp dtp dd
Jo Jo Jo
f2n r 1 / 7 \
= J / -(sin<p+2cos cp sirup) dcp d9
Jo Jo 5
-f
5 Jo
f
Jo
-lie
1 f2n 10
coscp — - cos <p
4n
dG = — .
5 7n 3 3
On the other hand
/ m = / / tBY(S)(Tft, TSz)dsids2
JSM J J[0,jr]x[0,2jr)
p2tz pTZ
= I I cos S\ dx A dy(TSl , TS2) ds\ ds2
Jo Jo
p2tz pit
Jo Jo
COS Si
cos S\ cos 52 — sin si sin s2
cos si sin S2 sin si cos S2
ds\ ds2
PZJV pit
= / cossi (cossi sinsi) ds\ d$2
Jo Jo
= / /
Jo Jo
cos si sin si Jsi ds2
p2n / y \ n /*2jr
2 An
- ds2 = — •
3 3
Therefore, the generalized Stokes's theorem is verified in this case.
8.3 I The Generalized Stokes's Theorem 557
Besides being notationally elegant, the integral formula in Theorem 3.2 beau-
tifully encompasses all three of the major results of vector analysis, as we now
show.
First, let co be a 1-form defined on an open set U in R2. Then
co = M(x, y)dx + N(x, y)dy,
so that
dco = dM A dx + dN A dy
dM J dM \ J (dN J dN J
dx H dy I A dx + I dx H dy ) A dy
dx dy J \ dx dy
dN dM
dx
dy
dx A dy.
The generalized Stokes's theorem (Theorem 3.2) says that if D is a 2-manifold
contained in U and dD is given the induced orientation (see Figure 8.9), then
/ dco = I co,
J D JdD
or, in this instance, that
* d \ dx
which is Green's theorem.
dN dM\
dxdy=(t Mdx + Ndy,
dy J JdD
= !L (f - f ) dx dy L°> = LM dx+Ndy
Figure 8.9 The generalized Stokes's theorem implies Green's theorem.
Next, suppose co is a 1-form defined on an open set U in R3. Then
co = F\(x, y, z)dx + F2(x, y,z)dy + F3(x, y, z)dz-
It follows that
,'9*3 3F2\ /9Fj 3F3\
flftj = I I ay A dz + I \ dz Adx
dy dz J \ 3z 3^
Chapter 8 | Vector Analysis in Higher Dimensions
Recall from Proposition 2.2 that if 5 is a parametrized 2-manifold (surface in R3),
then
/ co = (L }
JdS JdS
F-ds,
where F = F\ i + F2 j + F3 k. From Proposition 2.4,
/ dco = I I G-dS,
Js J Js
where
G=(^_^Wf---Vf---V=VxF.
\ dy dz ) \ dz dx J \ dx by J
Theorem 3.2 tells us, if S is oriented and 35 is given the induced orientation, that
or, equivalently, that
/ co = I dco,
Jas Js
(h F-dS= J J VxF-dS,
JdS J J S
which is the classical Stokes's theorem. (See Figure 8.10.)
J dco = JJ V x F ■ dS
Ls(o=LF-ds
Figure 8.10 The generalized Stokes's theorem gives the classical Stokes's
theorem.
Finally, let co be a 2-form defined on an open set in R3 . So
co = F\(x, y, z)dy A dz + F2(x, y, z)dz A dx + F3(x, y, z)dx A dy.
You can check that
SdFi dF2 dF3\
dco = I 1 1 I dx A dy A dz.
\ dx dy dz )
If D is a region in R3, then D is automatically a parametrized 3-manifold since
the map X: D —> R3, X(x, y, z) = (x, y, z) parametrizes D. (One can show that
in this instance D is always orientable as well.) If D is bounded and 3D (which is
a surface) is given the induced orientation (i.e., outward-pointing normal), then
8.3 I The Generalized Stokes's Theorem 559
Proposition 2.4 states that
f co = £ F-dS,
JdD JJ3D
where F = F\ i + Fi j + F3 k. From Example 6 of §8.2,
[ dco= f (— + — + — )dx Ady Adz= SIS V-FdV.
Jd JD\dx dy dz J J J JD
Theorem 3.2 indicates that fgD co = fD dco or
(jf F-dS = [ [ [ V-FdV,
JJdD J J Jd
which is, of course, Gauss's theorem. (See Figure 8.11.)
Figure 8.1 1 The generalized Stokes's theorem gives
rise to Gauss's theorem.
In the foregoing remarks, we have implicitly set up a sort of "dictionary"
between the language of differential forms and exterior derivatives and that of
scalar and vector fields. To be explicit, see the table of correspondences shown in
Figure 8.12.
The theorems of Green, Stokes, and Gauss all arise from Theorem 3.2
applied to 1 -forms and 2-forms. The next question is, can the "dictionary" and
Theorem 3.2 provide a corresponding result for 0-forms? The generalized Stokes's
theorem (Theorem 3.2) states, for a 0-form co and an oriented parametrized curve
C, that
/ dco = I co.
Jc Jbc
k
Differential &-form
Field
Derivative
0
Scalar field /
dco -0- V/
1
co = F\ dx + F2 dy + Ft, dz
Vector field
dco •*» V x F
F = Fx i + F2 j + F3 k
2
co = F\ dy A dz + Fi dz A dx + F3 dx A dy
Vector field
dco +± V • F
F = Fx i + F2 j + F3 k
Figure 8.1 2 A differential forms-vector fields dictionary.
Chapter 8 i Vector Analysis in Higher Dimensions
Now, if C is closed, then 3C is empty (and so fac co = 0). But if C is not closed,
then 3C consists of just two points. In that case, what should fgc a> mean? In
particular, to apply Theorem 3.2, we must orient dC in a manner that is consistent
with the orientation of C, which can be done by assigning a "— " sign to the initial
point A of C and a "+" sign to the terminal point B. (See Figure 8.13.) Then
fdC co is just f(B) — f(A), where / is the function (scalar field) corresponding
to co in the table. Since dco corresponds to V /, Theorem 3.2 tells us that
f Vf.di = f(B)-f(A), (1)
the result of Theorem 3.3 in Chapter 6.
Finally, for the case n = 1, that is, the case of 0-forms (functions) on R, the
0-form co corresponds to a function / of a single variable, and V/ is the ordinary
derivative /'. Furthermore, a parametrized curve in R is simply a closed interval
[a, b]. Then equation (1) reduces to
b
f(x)dx = f(b)-f(a),
a version of the fundamental theorem of calculus. Thus, we can appreciate that
the generalized Stokes's theorem is an elegant and powerful generalization of the
fundamental theorem of calculus to arbitrary dimensions.
8.3 Exercises
Figure 8.13 The orientation
of the curve C induces an
orientation of its boundary
(i.e., the endpoints A and B).
In Exercises 1-7, determine dco, where co is as indicated.
1.
CO =
exyz
2.
CO =
x3y — 2xz2 + xy2z
3.
CO =
(x2 + y2) dx + xy dy
4.
CO =
x\ dx2 — X2 dx\ + X1X4 dx4 — X4X5 dxs
5.
CO =
xz dx A dy — y2z dx A dz
6.
CO =
X1X2X3 dx2 A dxi A dx4 + X2JC3X4 dx\ A dx2
A
dxT,
7.
CO =
Yll=i xf dx\ A • • • A dxj A • • • A dxn (Note:
dxi
means that the term dxt is omitted.)
8. Let u be a unit vector and / a differentiable function.
Show that dfXo(u) = Duf (x0). (Recall that Du/(x0)
denotes the directional derivative of / at xo in the di-
rection of u.)
9. If co = F(x, z) dy + G(x, y) dz is a (differentiable) 1-
form on R3, what can F and G be so that dco =
z dx A dy + y dx A dz7
1 0. Verify the generalized Stokes's theorem (Theorem 3.2)
for the 3-manifold M of Exercise 11 of §8.2, where
&) = 2x dy A dz — Z dx A dy.
1 1 . Verify the generalized Stokes's theorem (Theorem 3.2)
for the 3-manifold
M={(x, y, z, ui)eR4| x=S - 2y2 - 2z2 - 2w2, x > 0}
and the 2-form co = xy dz A dw. (Hint: First compute
f3M co. To calculate fM dco, study Example 3 of this
section.)
12. (a) Let M be a parametrized 3-manifold in R3 (i.e., a
solid). Show that
1 f
Volume of M = — I x dy A dz — y dx A dz
3 Jsm
+ z dx A dy.
(b) Let M be a parametrized n -manifold in R". Ex-
plain why we should have
n -dimensional volume of M
1 f
= — I X\ dx2 A • • • A dx„
n Jsm
— X2 dx\ A dxi A • • ■ A dxn
+ Xi dx\ A dx2 A dX4 A • • • A dx„ + • • •
+ (— \)"~ xn dx\ A dx2 A • • • A dxn-\.
Miscellaneous Exercises for Chapter 8
True/False Exercises for Chapter 8
1. (dx Ady + dy Adz)((l,0, 1), (0,-1,3))= 0.
2. dx\ A d\2 A dxi A dx4 = dx2 A dx$ A dx\ A c/.^.
3. There are 21 basic 5-forms in R7.
4. rfxi A dx2 = dx2 A dx\.
5. (t/jti A </X2) A dx?, = dxi A (dx\ A c/-t2)-
6. If a) is a 3-form on R6 and r] is a 5-form on R6, then
d)M) = l)AO).
7. If a) is a 2-form on R8 and )? is a 3-form on R8, then
CO A Tj = T] A CO.
8. dx Ady A dz(a, b, c) = — c/z A dy A dx(a, c, b).
9. dxj a dxj(a, b) = —dx, a dxj(b, a).
10. Let D = [0, 2] x [-1, 1] and let X: D R4 be given
by
X0,0 = 0-f, Jf2, se',4t).
Then M = X(D) is a smooth parametrized 2-manifold
inR4.
11. Let D = [-2, 2] x [0, 5] x [-3, 3] and let X: D
R4 be given by
X(«i, M2, M3) = (u\ll\, «2COSM3' U\ — U2, u\u\).
Then M = X(£>) is a smooth parametrized 3 -manifold
inR4.
12. If D = [0, 1] x [0, 1], then the underlying manifolds
of X: D R3,
X(s, t) = (s coslirt, .ysin2jr?, s2)
andY:D -> R3,
Y(s, t) = (t cos27rj, fsin27rs, f2)
are the same.
13. Let co = dxAdy and D = [0, 1] x [0, 1]. Then
J a) = jY co, where X: D — > R3,
X(s, t) = (s coslirt, ssin2jtt, s2),
andY:D -* R3,
Y(s, t) = (tcoslirs, ?sin27rs, f2).
14. Let 5 = {u e R3 | u\ + u\ + u\ < 1}. The general-
ized paraboloid X: B R4 defined by
X(u\, U2, M3) = (u\, 142, M3, M j + 2l<2 + 3k2)
has as its boundary the ellipsoid Y: [0, jt] x [0, 27r) — >
Y(s, t) = (sins cost, A= sins sin t, -Lcoss, 1).
15. Let M C R" be the graph of a function f:UQ
R""1 -»■ R" parametrized by X: U -> R",
X(Mi, . . . , M„_i) = («!,..., M„_i, /(Mi, . . . , M„_i)).
If
TVT/ \ (/"I fun-1 ' ~ 1)
N(«l, . . . , =
y/;21+--- + /IL + 1
is a unit normal, then the parametrization X is compat-
ible with the (n — 1 )-form f2 defined by
^x(U)(ai , . . . , a„_i) = det [ ai • • • a„_i N ] .
16. If co = x-[Xi dx2 A dx4, then dco = X3 dx\ A dx2 A
dx4 + x\ dx2 A dxj A dx4.
17. If co = x\ dxi — X2 dx\ + x\X2Xt, dxi, then
dco = (X2X3 + \)dx\ A dxi + dx\ A dx2
+ X1JC3 dx2 A dx?,.
18. If co = x\X2 dx\ A dx2 + X2X3 dx\ A dx-$ + x^x? dx2
Adxi, then
dco = 2x2 dx\ A dx2 A dx$.
19. If co is an H-form on R", then dco = 0.
20. If M is a parametrized fc-manifold without boundary
in R" and co is (k — 1 )-form defined on an open set
containing M, then fM dco = 0.
Miscellaneous Exercises for Chapter 8
1. Let co be a £-form, an /-form. Show that
d(co A rj) = dco At] + (— l)kco A dr\.
This is accomplished by the following steps:
(a) Show that the result is true when k = I = 0, that is,
when co = f and ii = g. (Here / and g are scalar-
valued functions.)
(b) Establish the result when k = 0 and / > 0.
(c) Establish the result when k > 0 and / = 0.
(d) Establish the result when k and / are both positive.
2. Let M be the subset of R5 described as
{(x\, X2, Xt,, X4, X5) | X5 = X1X2X3X4, 0 < X\, X2, Xl,
x4 < 1}.
Chapter 8 | Vector Analysis in Higher Dimensions
(a) Give a parametrization for M (as a 4-manifold)
and check that your parametrization is compati-
ble with the orientation 4-form f2 = dx\ A dx2 A
dxi A dx$.
(b) Calculate JM X4 dx\ A <ijC2 A dx^ A J.«5.
3. (a) Let C be the curve in R2 given by y = f(x),
a < x <b. Assume that / is of class C1. If C
is oriented by the direction in which x increases,
show that if co = y dx, then
L
area under the graph of /.
(b) Let S be the surface in R3 given by the equation z =
f(x, y), where (x, y) e [a, b] x [c, d]. Assume
that / is of class C . If S is oriented by upward-
pointing normal, show that if co = z dx A dy, then
volume under the graph of / .
(c) Now we generalize parts (a) and (b) as follows:
Suppose f-.D—* R is a function of class C1 de-
fined on a connected region D c R"_1 . Let M be
the (n — l)-dimensional hypersurface in R" de-
fined by the equation x„ = f(x\, . . . , *„-i), where
(jci, . . . , G D. If co = x„ dx\ A • • • A dxn-\,
show that
JM
co = ±(«-dimensional volume
under the graph of /).
How can we guarantee a "+" sign in the equation?
4. Let M be the portion of the cylinder x2 + z2 = 1,
0 < y < 3, oriented by unit normal N = (x, 0, z).
(a) Use N to give an orientation 2-form £2 for M. Find
a parametrization for M compatible with Q.
(b) Identify 9M and parametrize it.
(c) Determine the orientation form Q3M for 3M in-
duced from n of part (a).
(d) Verify the generalized Stokes's theorem (Theorem
3.2) for M and co = zdx + (x + y + z)dy—x dz.
5. Use the generalized Stokes's theorem to calcu-
late fs4 co, where S4 denotes the unit 4-sphere
l(xi , x2, xi,X4, X5) G R5 I x\ + x\ + x\ + x\ + x\ =
1} and co = x^ dx\ A dx2 A dx4 A dxs + X4 dx\ A
dx2 A dxz A dxs .
6. (a) Let co be a 0-form (i.e., a function) of class C2.
Show that d(dco) = 0.
(b) Now suppose that is a &-form of class C2, mean-
ing that when co is written as
Pi\ ...it dxix A A dXjk ,
each Fi,.„i, is of class C2. Use part (a) and the result
of Exercise 1 to show that d(dco) = 0.
7. In this problem, show that the equation d(dco) = 0 im-
plies two well-known results about scalar and vector
fields.
(a) First, let co be a 0-form (of class C2 ). Then co corre-
sponds to a scalar field / . Use the chart on page 559
to interpret the equation d(dco) = 0.
(b) Next, suppose that co is a 1-form (again of class
C2). Then co corresponds to a vector field. Inter-
pret the equation d(dco) = 0 in this case.
8. Let
x dy A dz + y dz A dx + z dx A dy
(x2 + y2 + z2fl2
(a) Evaluate fs co, where S is the unit sphere x2 +
y2 + z2 = 1, oriented by outward normal.
(b) Calculate dco.
Figure 8.14 Figure for Exercise !
(c) Verify Theorem 3.2 over the region M =
{(x, y, z) \ a2 < x2 + y2 + z2 < 1}, where a / 0.
(d) Now let M be the solid unit ball x2 + y2 + z2 < 1 .
Does Theorem 3.2 hold for M and col Why or why
not?
(e) Suppose that S is any closed, bounded surface that
lies entirely outside the sphere S( = {(x, y, z) \
x2 + y2+z2 = e2}. (See Figure 8.14.) Argue that if
S is oriented by outward normal, then fs co = An .
9. Let M be an oriented (k + l + 1 )-manifold in R"; let
&) be a fc-form and r\ an /-form defined on an open set
of R" that contains M. If dM = 0, use Theorem 3.2
and Exercise 1 to show that
/ dco A 11
JM
(-1)
Ml
/ coAdrj.
J M
10. Let M be an oriented k -manifold. Use Exercise 1 and
the general version of Stokes's theorem to establish
"integration by parts" for &-forms co and 0-forms /:
l<ii <---<it <n
f fdco
J M
/
fto
I df A co.
JM
Suggestions for Further
Reading
General
Francis J. Flanigan and Jerry L. Kazdan, Calculus Two: Linear
and Nonlinear Functions, 2nd ed., Springer, 1990. The essen-
tials of vector calculus are presented from a linear-algebraic
perspective.
John H. Hubbard and Barbara Burke Hubbard, Vector Calculus,
Linear Algebra, and Differential Forms: A Unified Approach,
2nd ed., Prentice Hall, 2002. Treats the main topics of multi-
variable calculus, plus a significant amount of linear algebra.
More sophisticated in approach than the current book, using
differential forms to treat integration.
Jerrold E. Marsden and Anthony J. Tromba, Vector Calculus,
5th ed., W. H. Freeman, 2003. This text is probably the one most
similar to the current book in both coverage and approach,
using matrices and vectors to treat multivariable calculus
inR".
Jerrold E. Marsden, Anthony J. Tromba, and Alan Weinstein,
Basic Multivariable Calculus, Springer/W. H. Freeman, 1993.
Somewhat similar to Marsden and Tromba's Vector Calculus,
but with less emphasis on a rigorous development of the sub-
ject. A good guide to the main ideas.
George B. Thomas and Ross L. Finney, Calculus and Ana-
lytic Geometry, 9th ed., Addison-Wesley, 1996. A complete
treatment of the techniques of both single-variable and multi-
variable calculus. No linear algebra needed. A good reference
for the main methods of vector calculus of functions of two
and three variables.
Richard E. Williamson, Richard H. Crowell, and Hale F.
Trotter, Calculus of Vector Functions, 3rd ed., Prentice Hall,
1 972 . A smooth and careful treatment of multivariable calculus
and vector analysis, using linear algebra.
More Advanced Treatments
The following texts all offer relatively rigorous and theoretical
developments of the main results of multivariable calculus. As
such, they are especially useful for studying the foundations of
the subject.
Tom M. Apostol, Mathematical Analysis: A Modern Approach
to Advanced Calculus, 1st ed., Addison-Wesley, 1957. Look at
the first edition, since the second edition contains much less
regarding multivariable topics.
R. Creighton Buck, Advanced Calculus, 3rd ed., McGraw-Hill,
1978. Treats foundational issues in both single-variable and
multivariable calculus. Uses the notation of differential forms
for considering Green's, Stokes's, and Gauss's theorems.
Richard Courant and Fritz John, Introduction to Calculus and
Analysis, Vol. Two, Wiley-Interscience, 1974. A famous and
encyclopedic work on the analysis of functions of more than
one variable (Volume One treats functions of a single variable),
with fascinating examples.
Wilfred Kaplan, Advanced Calculus, 3rd ed., Addison-Wesley,
1984. A full treatment of advanced calculus of functions of two
and three variables, plus material on calculus of functions of n
variables (including some discussion of tensors). In addition,
there are chapters on infinite series, differential equations, and
functions of a complex variable.
O. D. Kellogg, Foundations of Potential Theory, originally pub-
lished by Springer, 1929. Reprinted by Dover Publications,
1954. A classic work that ventures well beyond the subject of
the current book. The writing style may seem somewhat old-
fashioned, but Kellogg includes details of certain arguments
that are difficult to find anywhere else.
James R. Munkres, Analysis on Manifolds, Addison-Wesley,
1991. A superbly well written and sophisticated treatment of
calculus in R" . Requires a knowledge of linear algebra. In-
cludes a full development of differential forms and exterior al-
gebra to treat integration. For advanced mathematics students.
David V Widder, Advanced Calculus, 2nd ed., originally pub-
lished by Prentice-Hall, 1961. Reprinted by Dover Publica-
tions, 1989. Careful treatment of differentiation and integra-
tion of functions of one and several variables. Chapters on
differential geometry, too.
Physics Oriented Texts
Mary L. Boas, Mathematical Methods in the Physical Sciences,
2nd ed., Wiley, 1983. A wide variety of topics that extend well
beyond vector calculus. For the student interested in physics as
well as mathematics.
Harry F. Davis and Arthur D. Snider, Introduction to Vector
Analysis, 6th ed., William C. Brown, 1991. A detailed and rel-
atively sophisticated treatment of multivariable topics. Includes
appendices on classical mechanics and electromagnetism.
563
564 Suggestions for Further Reading
Edward M. Purcell, Electricity and Magnetism, 2nd ed.,
McGraw-Hill, 1985. This is a physics, not a mathematics, text.
Provides excellent intuition regarding the meaning of line and
surface integrals, differential operations on vector and scalar
fields, and the significance of vector analysis.
Other
Alfred Gray, Modern Differential Geometry of Curves and Sur-
faces with Mathematica® , 2nd ed., CRC Press, 1998. Not
a vector calculus text by any means, but rather a delightful
library of geometric objects and how to understand them via
Mathematica. Some of the differential geometric topics are
somewhat remote from the subject of the current book, but the
many illustrations are worth viewing. An outstanding aid for
developing one's visualization skills.
H. M. Schey, Div, Grad, Curl, and All That: An Informal Text
on Vector Calculus, 3rd ed., W. W. Norton, 1997. The classic
"alternative" book on vector analysis, aimed at students of elec-
tricity and magnetism. Brief, but well done, intuitive account
of vector analysis.
Answers to Selected
Exercises
Chapter 1
Section 1.1
1. (a) y
.(2,1)
(b)
(3,3)
(c)
(-1,?)
3. (a) (2, 8)
(b) (-16,-24)
(c) (5, 15)
(d) (-19,25)
(e) (8, -26)
b (translated)
7. (a) AB = (-4,3,
(b) AC = (1, 1,3),
(-4,3,-1)
(c) In general, we have
1),BA = (4, -3, 1)
SC = (5, -2,4),
AC + CB
9. x = 14, v = 16, z = 8
1 1 . Because b = 5a.
13. (1, 2, 3, 4) + (5, -1,2, 0) = (6, 1, 5, 4)
2(7,6, -3,1) = (14, 12, -6, 2)
a + b = (fli + b\ , . . . , an + bn); kn = (ka\ , . . . , kan)
15. = (1,-4, -1,1)
17. If your displacement vector is a and your friend's is b, then
b — a is the displacement vector from you to your friend.
21. (b) The position vectors of points in the parallelogram
determined by (2, 2, 1) and (0, 3, 2)
23. (a) V5 units per minute
(b) (0, -4)
(c) 7 minutes
(d) No
25. (a) (5,5,4)
(b) (-5, -5, -4)
Fi = (4, -4, 2) and F2 = (-4, 4, 8)
27
Section 1.2
1. 2i + 4j
5. 2i + 4j
3. 3i + 7T j - 7k
7. (9, -2,V2)
565
566 Answers to Selected Exercises
9. (JT.-1)
11. (a) b = 2ai + a2
(b) b = — lai +4a2
(c) Take a = (bi + b2)/2, c2 = (&, - b2)/2.
13. a: = f + 2, j = 3f - 1, z = 5 - 6f
15. x = t + 2, y = -It - 1
17. x = t + 1, j = 4, z = 5 - 6f
19. ;ti = 1 - 2t, x2 = 5t + 2, x3 = 3f , x4 = 7t + 4
21 . (a) x=2t-l,y = 7-t,z = 5t + 3
(b) x = 5 - 5t, y = At - 3, z = 5t + 4
(c) One alternative for (a): x = —2t — l,y = t + 7,
z = 3 — 5t ; one alternative for (b): x = 5t, y = 1 —
At, z = 9 - 5t
(d) For (a): (x + l)/2 = (y - 7)/(-l) = (z - 3)/5;
for (b): (x - 5)/(-5) = (y + 3)/4 = (z - 4)/5;
for (c): (* + l)/(-2) = (y-7)/l = (z-3)/(-5)
and.x/5 = (y - l)/(-4) = (z - 9)/(-5).
23. (* - 7)/l = (y + 9)/3 = (z - 6)/(-8)
25. x = 3? - 5, y = 7f + 1, z = -2t - 10
27. Multiply the first symmetric form by — \, Then add — \
to each "side."
29. No. Setting t = 0 in the equations for l\ gives the point
(2, —7, 1). To have x = 2 in l2, we must take t =\. But
? = j gives the point (2, —7, 2) on l2, not (2, —7, 1).
31 . No, they do not. Note that x > — 1, y > 3, and z < 1
only. The parametric equations define a ray with endpoint
(-1,3,1).
33. (-14,5,-18)
35. With x = 0: (0, f , \); with y = 0: (-f , 0, with
z = 0: (7,17,0).
37. No 43. No
45. (a) Circle a2 + y2 = 4, traced once counterclockwise. If
0 < t < 2n, circle is traced three times.
(b) Circle x2 + y2 = 25, traced once counterclockwise.
(c) Circle x2 + y2 = 25, traced once clockwise.
(d) Ellipse *2/25 + y2/9 = 1, traced once counterclock-
wise.
47. x = (a + a6) cos 9, y = (a + a6) sinS
Section 1.3
1 . a • b = 13, ||a|| = ^26; ||b|| = 7T3
3. a • b = -44, ||a|| = V50; ||b|| = 2^14
5. a-b = 2, ||a|| = V26; ||b|| = ^3
7. 2;r/3 9. cos"1 (V2/V3") 11. n/2
13. |(i + j) 15. 2k
17. (2i-j + k)/V6
19. V3(i + j-k)
21 . Yes, if a • b = 0 or a = b.
25. (a) Work = (component of F)||Pgj|
= ||F|| cos6»|pe| = F-PQ
(b) 2
27. 10,000 ft-lb
29. cosa = 3/5, cos/3 = 0, cosy = 4/5
33. Hint: The diagonals are given by di = a + b and d2 =
b — a, where a and b determine the adjacent sides of the
parallelogram.
Section 1.4
1. 2 3. -5
5. (31,-5,8) 7. 5k
9. -6i+14j + 4k 11. 5730
13. c is parallel to the plane determined by a and b (or else a
is parallel to b).
15. V 1002/2
21. (axb)-c
17
3^
Z34/2
a\
0.2
a3
bi
b2
b3
C\
C2
C'3
bi
b2
b3
C\
C2
C'3
a\
0.2
«3
Expand determinants to see they are equal.
23. (b) 1
25. (a) a x b
(b) 2a x b/||a x b||
(e) a x (b x c)
(f) (a x b) x c
35. (a) 20V3 ft-lb
(b) 30V3 ft-lb
37. 75V3 ft-lb 39. 400;r/3 in/min
41 . Rotation about an axis that passes through the rigid body
is very different from rotation about a parallel axis that
does not pass through the body.
Section 1.5
1 . x - y + 2z = 8
3. 5x-4y + 3z = 25
5. 5x - Ay + z = 12
7. x-y + lz = 5
11. 3x - 2y - z = 3
Answers to Selected Exercises 567
13. x = 13f
15. -4/3
17. x=2s + t -
1 9. x = 2s + 5t
21. x = 3s + 3t
2i>y
-I4t+24,z
-5t
l,y = 2 - 3s, z = s - St + 7
- 1, y = lOt - 3s + 3, z = 4s + It - 2
- 5, y = 10 - 3s - 6t, z = 2s-2t + 9
23. I9x - I6y + lz
25. 31/V34
59
27. 25/V64T
(b) The lines must intersect.
5z = 2±3V35
29. (a) 0
31. 7T4/2
35. x + 3y
37. Hint: Consider Example 8 in §1.5 with A as Pi and B
as P2.
Section 1.6
1.
e„
1)
1)
(a) ei + 2e2 + 3e3 H 1- ne„
(b) ei - e3 + e4 - e6 H h e„_2 ■
(1,-2, 3,-4, ...,(-l)n+1n)
(a) (3,-1, ll,-l,...,2« + (-l)"+12n
(b) (-1, 7, -1, . . . , 2n + (-l)n2n - 1)
(c) (-3,-9,-15, ...,-6n + 3)
(d) %/l + 9 + 25 + --- + (2n- l)2
(e) 2 - 12 + 30 + • • • + (-l)"+12«(2n -
Hint: Use the triangle inequality.
Hint: Square both sides of the equation and write as dot
products.
1 3. Hyperplane in R5 passing through (1, —2, 0, 4, —1) and
perpendicular to the vector n = (2, 3, —7, 1, —5)
15. (a) Total cost =
(200, 250, 300, 375, 450, 500) • (xux2, x3,x4, x5, x6)
(b) {x e R6 | (200, . . . , 500) • (xu . . . , x6) < 100,000}.
Budget hyperplane is 200.x i + • • • + 500*6 = 100,000.
17.
9
11
19.
21. -42
5
8
-2
5
-4
0
15
-9
5
0
23. -240
25. (a) A lower triangular matrix is an n x n matrix whose
entries above the main diagonal are zero,
(b) For an upper triangular matrix, use cofactor expansion
about the first column or last row.
27. -289
31.
1/2
0
0
-1 1/2
1 0
0 -1
37.
14
14
J_
14
39. (32,46, 19,-35)
Section 1.7
1. (1,1)
5. (2^2, 3jt/4)
9. (-1/2, V3/2,-2)
13. (0,0,-2)
17. (2V2, jt/6, 7jt/4)
19. (a)
3. (3,0)
7. (2 cos 2, 2 sin 2, 2)
11. (o,3V3/2,3/2)
15. (2,2tt/3, 13)
d = nl2
(b)
21.
23. Cartesian: v = 2; cylindrical: r sin 6 = 2; surface is a ver-
tical plane.
25. Cartesian: x = y = 0; spherical: <p = 0 or jt; object is the
z-axis.
35. Note: The desired region is that between the two spheres
shown.
z
y
37. (a) r = —f(0) is the reflection through the origin of
r = f(0).
(b) p = — f{<p, 9) is the reflection through the origin of
p = f(cp,9).
(c) r = 3f(9) is a threefold magnification of r = f{6).
(d) p = 3f(<p,6) is a threefold magnification of
p = f(cp,9).
41 . i = sin<p cos# ep + cos <p cos 9 ev — sin 9 eg
j = sin <p sin 9 ep + cos cp sin 9 ev + cos 9 eg
k = cos (pep — sin 0
45. (a) Hint: use part (a) of the previous exercise.
47. Hint: use part (a) of the previous exercise.
True/False Exercises for Chapter 1
1 . False. (The corresponding components must be equal.)
3. False. ((—4, —3, —3) is the displacement vector from P2
to Pi.)
5. False. (Velocity is a vector, but speed is a scalar.)
7. False. (The particle will be at (2, -1) + 2(1, 3) = (4, 5).)
9. False. (From the parametric equations, we may read a vec-
tor parallel to the line to be (—2, 4, 0). This vector is not
parallel to (-2, 4, 7).)
1 1 . False. (The line has symmetric form
13. False. (The parametric equations describe a semicircle
because of the restriction on t.)
15. False. (||*a|| = |&|||a||.)
1 7. False. (Let a = b = i, and c = j.)
19. True
21 . True. (Check that each point satisfies the equation.)
23. False. (The product BA is not defined.)
25. False. (det(2A) = 2" det A.)
27. False. (The surface with equation p = 4 cos cp is a sphere.)
29. True
Answers to Selected Exercises 569
Miscellaneous Exercises for Chapter 1
3. x = 63t + 1, y = 148f, z = 847r - 2
5. (a) x = 5t + 2, y = 3t - 2
(b) x = (b2 - a2)t + (en + bi)/2,
(a2 + b2)/2
7. (a) 2xi + 4x2 — 2x3 + X4 — x5 = 5
(b) (bi - a\)x\ -\ \-(b„- a„)x„
y = (a\ - bi)t +
■a? +
9. (a) No
11. (a) tt/3
(b) x = t, y
(b) No
\-t,z = t
13. Planes (a) and (d) are parallel; (b) and (e) are the same;
(c) is perpendicular to (b).
1 5. The dot product measures the agreement of the answers.
17. Hint: a x b is normal to the plane determined by a
and b.
19. (a) 5^21 (b) 10
21. (b) 1
2 5 . Hint : Note that to determine x, it suffices to determine ||x||
and the angle between a and x. Consider the cases c = 0
and c / 0 separately.
29. (a) Hint: Let a and b denote two adjacent sides of the par-
allelogram.
(b) Hint: You should simply give a vector equation.
31. (b)A"
1 n
0 1
33. (a) H6
6 7 8 9 10
J_
10
11
1 1
det#2 = — , detff) = ,
12 2160
1 1
det#4 = , det//5 = ,
6048000 266716800000
1
det H6 =
186313420339200000
(b) det #10 « 2.16418 x 10"53
(c) Most likely, AB and BA will not equal
Sa + b
35. x = (a + b)cost — bcos ( — ; — t
,'a + b
y = (a + b) sin t — b sin ( — ; — t
' a — b
37. Hypotrochoid: x = (a — £>)cos/ + ccos | — : — t
' a — b
y = (a — b) sin t — c sin — ; — t \ ;
'a + b
Epitrochoid: x = (a + b) cos t — c cos | — : — t
,'a + b
y = (a + b) sin t — c sin | — : — t
39. (a)
-1.00
-1.00-0.75-0.50-0.25 0 0.25 0.50 0.75 1.00
(b)
570 Answers to Selected Exercises
Answers to Selected Exercises 571
(c) z
(d) z
45. (a) {(r, 6, z) | 0 < r < 3, 0 < 6 < 2n, 0 < z < 3}
if the cylinder is positioned so that the center of the
bottom is at the origin and its axis is the z-axis.
(b) Using the same positioning as in part (a), {(p, cp, 6) \
0 < p < 3 sec cp, 0 < (p < jt/4, 0 < 6 < 2n} U
{(p, <p, 6) | 0 < p < 3csc?), 7r/4 < <p < jt/2, 0 <
e < in}.
Chapter 2
Section 2.1
1 . (a) Domain = R; range = {>> | y > 1}
(b) No
(c) No
3. Domain = [(x, y) \ y / 0}; range = R
5. Domain = R3; range = [0, oo)
7. Domain = {(x, y) \ y ^ 1};
range = {(x, y, z) | y / 0, y2z = (xy - y - l)2
+ 0+1)2}
9. y, z, t) = xyzt,
v2(x, y, z, t) = x2 - y2,
Vi(x, y, z, t) = 3z + t
11. (a) f(x) = -2x/W
(b) Mx, y, z) = -2x/Jx2 + y2 + z2,
f2(x, y, z) = -2y/Jx2 + y2 + z2,
Mx,y,z) = -2z/Jx2 + y2 + z2
13. (a) fi(\) = 2xi - *3 - x4,
/2(x) = 3x2,
/3(x) = 2xi -X]-x4
(b) Range = {(yu y2, y3)\yi = >'3}
/ — y
A
17. y
X
-10 -5 0 5 10
Z
x
Answers to Selected Exercises 573
574 Answers to Selected Exercises
45.
Section 2.2
1 . Open
5. Neither
9. Does not exist
13. 0 15. 0
3. Neither
7. 2
1 1 . Does not exist
17. 0 19. -1
Answers to Selected Exercises 575
21. Limit does not exist.
23. Limit does not exist.
25. Limit does not exist.
27. Limit is 0.
11
9. fx(x, y) = ey + 2xy cos (x2 + y),
fy(x, y) = xey + sm(x2 + y) + y cos(x2 + y)
Fz = V(y + z),
F-
: -(x + z)/{y + zf,
(y-*)/(y + z)2
13. Fx = x/^Jx2 + y2 + z2,
Fy = y/Jx2 + y2 + z2,
Fz = z/^Jx2 + y2 + z2
15. F
1 - 2x2 - 3xy + y2 - 3xz + z2
Fy
17. Fx
Fy
(l+x2+y2 + z2)V2
1 + x2 - 3xy - 2y2 - 3yz + z2
{\ + x2 + y2 + z2?'2
1 + x2 + y2 - 3xz - 3yz - 2z2
(1 + x2 + y2 + z2)5'2
x4 - 2xyz + 3x2z2 + 3x2
(x2 + z2+ l)2
19.
21.
23.
F-
x2 + z2 + r
x2y — 2xiz — yz2 + y
(x2 + z2 + l)2
-i + (27T + l)j
+ j-2k
2k
25. i
29.
Limit does not exist.
27. [-4 -3/e
-3/e]
31.
2
29.
0 -2
0
33.
Limit does not exist.
_ 1/V5 0
-2/V5
35.
0
_
37.
Limit does not exist.
" 3 -7
1
0 "
39.
Continuous
31.
5 0
2
-8
41.
Continuous
_ 0 1 -
17
3 _
43.
Not continuous at (0, 0)
" -2 0 "
45.
Continuous
33.
1 -1
47.
Hint: Write /(x) in terms of the components of x.
0 2
Section 2.3
1 . fx(x, y) = y2 + 2xy, fy(x, y) = 2xy + x2
3. fx(x, y) = y cosxy — y sinxy,
fy(x, y) = x cosxy — x sinjcy
5. fx(x,y) = 4xy2/(x2 + y2)2,
fy(x, y) = -4x2y/(x2 + y2)2
7. fx(x, y) = — 3x2y sinx3y,
fy(x, y) = —x3 sinx3y
37. (a) The function has continuous partial derivatives.
(b) z = 3x + Sy + 3
39. z = e(x + y)
41 . x5 = -4x\ + 6x2 - 4x3 - 6x4 + 28
43. (a) h(0.l, -0.1)= 1
(b) /(0.1,-0.1)= 1
45. (a) fe(1.01, 1.95, 2.2) = 21.76
(b) /(1.01, 1.95, 2.2) = 21.6657
47. (a) fx(x,y) = (3x4 + Sx2y2 + y4)/(x2 + y2)2,
f (x, y) = -{x4 + 4x2y + 2x2y2 + y4)/(x2 + y2)2
(b) fx(0, 0) = 3, /,(0, 0)
■1
576 Answers to Selected Exercises
49. (a) (8 + A4l)x - 8y = V2(n - 4)
(b)
0 0.5
51. (a) x + y = ln2- 1
(b)
53. (a) z=Ux- 15
(b)
2.50
0.50
(c) Partial derivatives are polynomials — hence contin-
uous— so / is differentiable at (2, 1).
(c) Partial derivatives are rational functions defined on
R2 — hence, continuous — so / is differentiable at
(0, 0).
57. (a) z = -j^[(9-j3n - 96V2> - (16V2tt + 72)y +
(4V2 - 3V3)tt2 + (1672 + 9)tt]
(b)
(c) Partial derivatives are products of polynomials and
sine and cosine functions and, hence, are continuous.
Thus, / is differentiable at (tt/3, Jr/4).
59. Df(x) = A. Note that f'(x) = a in the one-variable case.
Section 2.4
1- D(f + g)
= [y — sinx + y cos(xy) x + 3y2 + x cos(xy)]
3. D(f+g)
siny + 3x2 cos x — x3 sinx xcosy 1
—6x + yz ez+xz yez + xy
5. D(/g)=[3x2 + y2 2xy],
D(f/g)=[y2-y4/x2 2xy + 4y3/x]
7. D(fg)
= [l2x3y + 3x2y5 - 12xy3 - 2y7
3x4 + 5x3y4 - 18x2y2 - 14xy6],
D{f/g)
~y(2y6 - 6x3 - 3x2y4) 3x3 + 6xy2 + 5x2y4 - 6y6
x2(x2 - 2y2)2 x(2y2 - x2)2 ,
9. /„ = 6xy\ f„ = 42x3y5 + 6x,
fxy /,, 2\vy< ■ 6 v 7
11. fx.
fx.
13. /,
-ye~x + 2yx-ie>'x + ylx~"eyix , fyy
fyx = e~
2[sin2 2x — cos 2x(sin2 x + 2ey)]
(sin2 x + 2ey)3
fyy =
15. /„ =
siny
2ey(2ey - silt x)
(sin2x + 2ey)3
fxy — fyx
4ey sin 2x
(sin2x +2ey)3
-y sinx, fyy = x cosy, fxy = fyx = cosx +
Answers to Selected Exercises
577
17. /„ = 2e\ fyy = x2e>, fzz = Ae2\ fxy = fyx = 2xe> ',
/,, ./:-. 0, j\: ./., 0
1 9. fxx = 2yz, fyy = 2xz, fa = 2xy, fxy = fyx = 2xz +
2yz + z2, fxz = fzx = 2xy + y2 + 2yz, fyz = fzy =
x2 + 2xy + 2xz
21 . fxx = b2ebx cosz + a2eax siny, fyy = —eax sin y, fzz =
-ebx cos z, fxy = fyx = aeax cos y, fxz = fzx =
-bebxsmz,fyz = fzy = 0
23. d"f/dx" = 3"yelx, d"f/dy" = 0, forn > 2
25. d"f/dx" = {-\)n-\n - \y./x",d"f/dy" = (-l)"-\n -
l)!/y", d"f/dz" = (-1)"(« - l)l/zn. All mixed partials
are zero.
27. (a) px and py have degree 16; pxx, pyy, and pxy all have
degree 15.
(b) px and py have degree 3; pxx has degree 2; pyy has
undefined degree; pxy has degree 2.
(c) The degree of dkp/dxil ■ ■ ■ dxik is d — k, where d
is the highest degree of a term of p of the form
xf'x* • • ■ xft" such that, for j = 1, 2, . . . , n, dj is at
least the number of times Xj occurs in the partial
derivative. If p has no such term, then the degree of
dkp/dxil ■ ■ ■ dxit is undefined.
29. (a)
T 0
(b)
Z 0
33. (a)
Section 2.5
1. df/dt = (18f2 + 14f)sin2? - 6 cos 2t sin2 2t + (12r3 +
14r2)cos2r + 72r + 84
2. df/ds = (3s2 + t2+ 2st) cos(0s + t)(s2 + 12)),
df/dt = (s2 + 3t2 + 2st)cos((s + t)(s2 + t2))
3. (a) dP/dt = (36 - 277r)/V2 atm/min
(c) Approximately (.36 + (9 - .27)tt)/V2
« 19.6477 atm
5. —6 in3/min. Decreasing.
7. 0.766 units/month
9. 0.2244 cm/min
13. Hint: dw/du = 2u(dw/dx) — 2u(dw /By) from the chain
rule.
15. Hint: dw/dx = (dw/du)({y^ - x2y)/(x2 + y2)2) from
the chain rule.
17. Hint: dw/dx = —\/x2(dw/du + dw/dv) from the chain
rule.
19. D(fog)= 2e2j_14/ _14^-i4f
578 Answers to Selected Exercises
21. D(fog)=[(l+s + t)es-' (\-s-t)es->]
3s2 -t2 - 1st
6s5P-s-2r2 3s6t2 -25-1/-3
23. D(fog)
25. D(fog):
27. D(fog)
e2,{2 sinf + cosf)
e'(sin2 t + 2 sin t cos f)
3(sin2rcosr + e3')
29. (a)
(b)
31. (a)
t + u
3s2f3 - tu2es'"2
7 10
31 44
S + M
3*3r2
1 13
2 31
dx2
drL r
sin 6 3 2sin# cosf? 3
3r
3r36»
2 sinfl cos 9 3 sin2 6» 32
+ = — +
iil)
r2 d92'
By2
sin2 6
dr2
2 cos26 3 2sin#cos# 32
+ — H
3r
3r36>
2 sin 6 cos 0 3 cos2# 32
+
33. (a)
dr
r2 36
3 cos <p 3
sin <p 1
dp p dtp
r2 d62
35. dy /dx = (2xy1 — y cos xy)/(x cosjty — lx2y6 + ey)
37. dz/dx = 3x2z3/(yz2 sinz + siny — x3z2), dz/dy =
(z2 cos z + z cos y)/(y z2 sin z + sin y — jc3z2)
' dw\ ( dw\ ( dw^
39. (a)
_ i (dw\ - 7
dx J y z ' \ dy Jx.z ' V 3z
. „ / dw\ / dw
1 - 20y
(b)
dw
~dx
(dw
~dx
' dw\ / dy\ ( dw
dy Jx.z \ dx J \ dz
3z\
41.
3 s
dz
(l)(l) + (7)(0) + (-10)(2*)
= xw — 2z;
xw — 2z +
2 3 3
x y — x
xw — 3y2z
Section 2.6
1. (a) The directional derivative of / at (x, y, z) in the
negative z-direction.
(b) V/(*,y,z)-(-k) = -3//3z
3. 8V5/5
5. (2e- 9)75/5
7. -4V3e-14
9. (a) fx(0, 0) = /,(0, 0) = 0
(b) -Dv/(0, 0) = v\w\ for all unit vectors v = v i + w j.
(c)
11. (a) (— 2i + j)/V5 direction
(b) ±(i + 2j)/V5 direction
13.
1 5. Travel along y3 = 27 x (toward the origin)
17. y-z = \ 19. x-2y-2z = 2
21 . The tangent plane has equation x — nz = 2.
23. (|, -£,_§) „d (-f.s, 9)
27. (a) 3jc+12v + 2z+10 = 0
(b) There is no tangent plane at (0, 0, 0).
29. Tangent line is 5x — 3^/4 y = — 1.
31 . Parametric equations: x = 5t + 5, y = 4t — 4. Cartesian
equation: Ax — 5y = 40.
33. Parametric equations: x = 14f + 2, y = t — 1. Cartesian
equation: x — 14y = 16.
Answers to Selected Exercises 579
35. x = -2t - 1, y = t - 1, z = t - 1
37. Xs — X\ = 71
39. xi + x% H 1- xB_i — x„ = V«
41 . (a) Near all points of S such that x / 0. At such points
f(x, y) = ln(l — sinxy — x3y)/x.
(b) All points (0, y, z) (i.e., the yz-plane).
(c)
43. (b) Yes, y = x2?\
y
1 -
0.5-
-0.5
0
0.5
X
(c) The implicit function theorem suggests that we need
not be able to solve for y in terms of x . Note that we
can in this case, but that the function fails to be of
class C1 at x = 0.
45. (a) G(-l, 1, 1)= F(-2, 1) = 0
(b) Hint: Use the chain rule to find Gz(-l, 1,1) =
30/0.
47. (a) A(l, 0,-1, 1,2) = -120/0; apply the general im-
plicit function theorem.
(b) ^(1, 0) = -I j^(l, 0) = -|, ^(1, 0) = -1
oxi oxi axi
49. (a) Anywhere except where <p = 0 or it .
(b) We can solve anywhere except along the z-axis, which
is where 6 can have any value.
Section 2.7
1. (1.302942538,-0.902880451)
3 (a) Estimate intersection points near (1, 1/2) and
(-1/2, -3/4).
y
1.5
-
1.0
0.5
-1.J
-1.0
-0.5
-0.5
0.5
i
-1.0
-1,5
(b) (1.103931711,0.441965716) and (-0.518214315,
-0.657923613)
5 (a) x0 = (-1, 1) leads to (-1.26491 11, 1.54919334).
(b) x0 = (l,-l) leads to (1.26491106,-1.5491933),
while x0 = (-l,-l) leads to (-1.2649111,
-1.5491933).
(c) It appears that an initial vector leads to an intersection
point that lies in the same quadrant.
9. (-0.9070154, -0.9070154)
True/False Exercises for Chapter 2
1 . False
3. False. (The range also requires v / 0.)
5. True
7. False. (The graph of x2 + y2 + z2 = 0 is a single point.)
9. False
1 1 . False, (lim^^-^o.o) f(x, y) = 0 / 2.)
13. False
1 5. False. (V/(x, y, z) = (0, cos y, 0).)
17. True
1 9. False. (The partial derivatives must be continuous.)
21. False. (./,, .- /,,.)
23. True. (Write the chain rule for this situation.)
25. False. (The correct equation is x + y + 2z = 2.)
27. True 29. False
580 Answers to Selected Exercises
Miscellaneous Exercises for Chapter 2
1 .(a) /i(x) = -x2, fi(s) = x\- *3, /3(x) = x2
(b) Domain is R3 ; range consists of all vectors in R3 of the
form a i + b j — a k.
3. (a) Domain: {(x, y) \x > 0, y > 0} U {(x, y) | x < 0,
y <0}
Range: [0, oo)
(b) Domain is closed.
5.
f(x,y) =
Graph
Level curves
1
x2 + y2 + 1
D
d
sin-y/x2 + y2
B
e
(3y2 - 2x2)e-x2-2y2
A
b
y3 — 3x2y
E
c
x2y2e-S-2f
F
a
-v2-v2
ye x }
C
f
7. 0
11. (a) 15°F
(b) 5°F
13. 1.3
15. (a) 9.57°F; 16.87°F; 8.39mph
(b) Effect of windspeed is greater in the Siple formula
than in the table.
(c) If t < 91.4°F or s > 4 mph, the Siple formula gives
windchill values that are higher than the air tempera-
ture, which is unrealistic.
120
(b)
w (io, o
20
, , yf^r , , i
-40 -20
^20
40
-40
-60
(c)
Wi
Answers to Selected Exerc'
Wn
1 9. Hint: Compare the normal vector to the tangent plane with
the vector O P .
21. (a) 3* - 4y - 5z = 0
(b) ax + by — cz = 0
23. (a) 2x - 8y - z = 5
(b) x=l,y = t-l,z = 5-8t
25. 12, 201.4 units/month
27. (a) dw/dp = 2p, dw/dip = dw/98 = 0
29. (a) dz/dr = e'(cos 6(dz/dx) + sin 0(dz/dy)),
dz/96 = er(- sme(dz/dx) + cos6(dz/dy));
dz/dx = e-r(cos6(dz/dr) - sm0(dz/d0)),
dz/dy = e-r{sm6(dz/dr) + cos8(dz/d6))
(b) Hint: d2z/dx2 = e-2r(cos2 6(d2z/dr2)
+ (sin20 - cos2 6)(dz/dr)
+ 2cos6»sin6»(9z/96»)
-2cos6»sin6»(92z/9r90)
+ sin26»(92z/902)),
-2'(sin26l(92z/9r2)
d2z/dy2
+ (cos26» - sin26»)(9z/9r)
-2sin6»cos6»(9z/96»)
+ 2sin6»cos6»(92z/9r96l)
+ cos2 6(d2z/d62))
31. u" [u{""-l) + uu" Inu + u""(\nu)2]
35. (a) z = f{x, y), where fix, y) = (x3 - 3xy2)/(x2 +
y2) ifix, y) / (0, 0); fix, y) = 0 if (*, y) = (0, 0)
(b) Yes
(c) df/dx = (x4 + 6x2y2 - 3y4)/(x2 + y2)2,
df/dy = -8x3y/(x2 + y2)2 (if (x, y) / (0, 0));
MO, 0) = i, fyio, o) = o
(d) £>u/(0, 0) = (3//9r)Uo = cos 39
(e) /y(0, 0) = 0 from (c), but along y = x,
fyix, x) = —2, which does not have a limit of 0.
37. Homogeneous of degree 3
39. Homogeneous of degree 3
41 . Homogeneous of degree 0
Chapter 3
Section 3.1
7. \(t) = 3i + 2j, speed = Vl3, a(f) = 0
582 Answers to Selected Exercises
9. v(f) = (sin? + t cost, cost — t sin?, 2t),
speed = V5f2 + 1,
a(f ) = (2 cos t — t sin t , —2 sin t — t cost, 2)
11. (a)
15. I(f) = fi + (3/ + l)j
17. 1(f) = {At - 4, \2t - 16, 80r - 128)
19. (a) y
(b) 1(f) = (t, 10? - 15)
(c) y =x3 -2x+\
(d) y — 5 = 10(x — 2). (Let x = t to verify agreement.)
21. 117.1875 ft
23. 25.09° or 64.91°
25. No
27. Write out both sides in terms of component functions.
29. Hint: Differentiate ||x(r)||2.
33. (a)x(l) = x(-l) = (l,0)
(b) jt/2
Answers to Selected Exercises 583
Section 3.2
0.5 1 1.5 2 2.5 3
(b) The path consists of three C 1 pieces: one for —2 < t <
0, another for 0 < t < 1, and another for 1 < t < 2.
(c) 4V2
1 5. Hint: Begin with x = r cos 6, y = r sin£>.
(-5 sin 3f i + 2j + 5 cos 3f k),
17. T:
N :
B
K -
19. T:
N :
B
K -
29
cos3ri
sin3f k,
1
(-2sin3f i- 5j + 2cos3fk),
29' 1
2
3
' 24
(1, ±Vf+T, -±VT=f),
2V2(o, iyr^r, ivm"),
■^(l,-Vt + T,VT=t),
V2/(67T^72), r = 1/ (sVT^T2 )
21. (b) K
23. (a)
sin x
ab
25. (a)
(a2 sin t + b2 cos2 tfl1
25
20
15
10
5
0
584 Answers to Selected Exercises
27. atang = 4f/V4f2 + 1, anorm = 2/V4f2 + 1
29. atang = V5e', flnorm = 2a/5V
31 . atang = 4f/V2 + 4f2, anorm = 2/Vl +2f2
33. (b) 0^ = 4f/V4f2 + 10, anorm = 27T0/V4f2 + 10
35. Hint: Write v and a in terms of T and N and use the
Frenet-Serret formulas.
37. Hint: Use formula (17) and Exercise 35.
39. Hint: Recall that ||x — xo||2 = (x — xo) • (x — xo).
41 . Hint: Calculate T and show that (x(r ) - (1 , 0, 0)) • T = 0.
Section 3.3
1. y
9.
/ / / / - -
/ / / y ' -
///'■■-
V V \ \
////,.
III!..
Ill
v \ V \
» > \ V
\ \ \ V x -
\ \ \ V x «
i i i i i
' ' / /
' ' / 1
\ \ N V - -
' ' / /
\\ N - -
' s / /
-
/ / / /
////
/// /
/ 1
\ \
\ W\
// / f
/ I
» \
\ \ w
/ / / '
/ i
\ \
/ s / /
/ /
S \ X X
>k V v s
N V \ \
\ \
/ /
✓ y s s
/ / / S
\\ \ \
\ V
/ /
/ / / /
\\ \ \
\ \
; /
/ / //
\\\ \
\ \
/ /
/ ///
11.
13.
. . - X "V \
\ \ \ \
\ v \ \
y \ \ t
1 t t
t t f
t f /
„_-.---•• s
V \ t t
f f /
t I 1
1 f t
v i t »
k 1 i y
k 1 ' ' '
- t J
/ / /
4 t /
///''''
///>'''•
/ / / t f ' '
/ i t * ♦ ' '
/ M 1 ' n
J ( M V * >
x x ^ ^
1 1 \ \ \ v \
Answers to Selected Exercises 585
15.
Section 3.4
■ • i
4 4 *■ »
f t ' '
t 1 '
~ 4 *
~ 4 4
- < <
' ' t
17. F(x(f)) = (cosf, -sinf,0) = x'(0
19. F(x(f)) = (cosf, -sin?,2e2') = x'(0
21.
23.
25.
31.
1
2-?
(a) F = V/, where f(x, y, z) = 3x — 2y + z.
(b) Equipotential surfaces are parallel planes with equa-
tion 3x — 2y + z = c (i.e., planes with normal
(3, -2, 1)).
Hint: Use the chain rule and the facts that V/ = F and x
is a flow line of F.
Hint: Differentiate the defining differential equation for
the flow with respect to x = (*i, X2, ■ ■ ■ , x„).
1
3
5
7
9
11
13
2yz)k
div F < 0
div F < 0 on
2x + 2y
3
2x\ + 4x2 + 6x3 + • ■ ■ + 2nxn
2xz i — 2yz j — ey k
0
(x2 - xyexyz)i + (y2 - 2xy)\ + (yzexyz
(a) div F > 0 on all of R2
(b) divF < 0 on all of R2
(c) div F > 0 on {(x, y)\x > 0},
{(*, y) \x < 0}
(d) div F > 0 on {(x, y) | v < 0},
{(*, y) I y > 0}
21 . Write out in terms of the component functions of F and G.
23. Write out in terms of the component functions of F.
25. Write out in terms of the component functions of F and G.
27. First use the chain rule to replace occurrences of the
Cartesian differential operators in V by combinations
of spherical differential operators. Then compute V/.
Finally, replace i, j, and k by appropriate combinations
of ep, e^, and eg. (See also §1.7.)
29. Write out the components of /V/.
33. (1/V3.4/V3.-1/V3)
True/False Exercises for Chapter 3
1 . True
3. True
5. False. (There should be a negative sign in the second term
on the right.)
7. True
9. False
1 1 . True
13. True
1 5. False. (It's a scalar field.)
17. True
1 9. False. (It's a meaningless expression.)
21 . True. (Check that F(x(0) = x'(O-)
23. False. (V x F / 0.)
25. False. (Consider F = y i + x j.)
27. True
29. False. (V • (V x F) / 0.)
Miscellaneous Exercises for Chapter 3
1. (a) D (b) F (c) A
(d) B (e) C (f) E
3. Hint: Differentiate (ds/dt)2 =
x'(0
586 Answers to Selected Exercises
w = 0
9. (a) Hint: Calculate x(0) and x(l).
(b) Hint: Show that
11 ■ (a) mwj^ ~ 2x2 + -*3)2 + (vi - 2y2 + y3f
(b)
X]
2x2 + x3)2 + (yi -2y2 + y3)2
2(l+u.)^
13. (a) Hint: Find where x'(f) = (0, 0).
(b) Hint: The tangent line at x(to) is given by \(s) :
x(?o) + Jx'(?o). Find the y-intercept of this line.
15. Hint: Begin with x(6>) = (/(0)cos6>, f(6)sm0).
17. (a) y(f) = (a(cos f + t sinr), a(sinf — t cos f))
(b) y
4
19. (a) Show that ||x(r) - y(r)|| =
(b) The vector difference x(f) — y(t) is a tangent vector
(to x) s(t) units long.
23. e(f) = (a(? + sin;), a(cosf — 1)), which is a another type
of cycloid.
25. Hint: Use the Frenet-Serret formulas.
27. (a) Show that (x'(s))2 + (y'(s))2 = 1 by means of the
fundamental theorem of calculus.
(b) K = \g'{s)\
(c) Set g'(s) = k(s). Find g by antidifferentiation and set
x(s) = So cos g(t)dt,y(s) = So sin g(t)dt.
(d) x(s) = f* cos(t2/2)dt, y(s) = f°sm(t2/2)dt
(e)
29. (a) Calculate B and remember that it is a unit vector,
(b) B is constant, so B' = 0. Next use the Frenet-Serret
formula.
a2 + h2/4jt2
31. (a) (b) 8.2771 ft
a
35. N' = -kT + tB. Argue that N' / 0.
37. (a) 242tx (b) F = V2(y i - 2x j + y k)
41 . Hint: Use Proposition 1.4.
43. No
Chapter 4
D3 +
9)3/3888
Section 4.1
p4(x) =l+2x + 2x2 + 4x^/3 + 2x4/3
p4(x) = 1 - 2(x - 1) + 3(x - l)2 - 4(x
5(* - l)4
P3(x) = 3 + (x - 9)/6 - (x - 9)2/216 + (x
Ps{x) =\-{x- nil)2 12 + (x - 7r/2)4/24
Pi{x, y)=\-2{x- l)/9 + 2(y + l)/9,
Pz(x, y)=\-2(x- l)/9 + 2(y + l)/9 + (x - l)2/27
- 8(x - l)(y + l)/27 + (y + l)2/27
Pi(x, y) = — 1 — 2x,
p2(x, y) = -l- 2x - 2x2 + 9(y - tt)2/2
1.
3.
5.
7.
9.
11
13. pi{x, y, z) ■
3y2 + 2xz
15. p\(x, y, z) -
1 + x + Sy + 4z
p2(x, y,z) = 0
p2(x, y, z) = xy
17.
VI
4
4
v2
4
4 -I
Answers to Selected Exercises 587
19.
6 2 0
2 0-2
0 -2 12
21. P2(*,y) = l + [ 0 0 ]
23. p2(x,y,z) = 2+[ 0 5 1]
2 0
0 -2
z-2
" 0
3
0
X
jc y z — 2 ]
3
8
2
y
0
2
0
_ z-2 _
25. (a) Df(0)=[ 1 2 ••• n ],
" 1 2 3 •••
ff/(0)
2 4 6
3 6 9
2n
3/7
« 2h 3« • • • «2
(b) Pi(*i, . . . , Xn) = 1 + Jti + 2x2 H 1" «*n
p2(*l, . . . , A„) = 1 + £"=1 «i + \ E" ,=i ?7*<*/
27. (a) 2 - z + 3xy + x2y - xz2 + 2y3
(b) -4 - 4(x - 1) + ll(y + 1) - z + |[4(jc - l)2 +
16(* - l)(y + 1) - 12(y + l)2 - 2z2] +
1[18(jc - l)3 + 24(x - l)2(y + 1) - 6(x - l)z2 +
12(y + l)3]
29. 2x dx + 6y dy — 6z2dz
31. ex cos j + (e-v sin z — eA' sin y)dy + ey cos z dz
33. (a) 388.08 (b) 0.24625 (c) 1.1
35. The ( 1 , l)-entry (upper left)
37. 2.4 cm
39. 0.0068 m
41. (a) P2(x,y)=\-\[x2+{y-«1)2}
(b) Accurate to at least 0.0360
43. (a) p2(a.,y)=|_y_2x(y-f)
(b) Accurate to at least 0.03 1 1
Section 4.2
1. (a) (2,3)
(b) Af = -h2 - k2
(c) There is a local maximum at (2, 3).
3. Local maximum at (|, |)
5. Local minimum at (y, 5); saddle point at (|, l)
7. Minimum at (4, |)
9. Saddle point at (0, 0); minimum at (0, 2)
1 1 . Saddle point at (0, ^); local minimum at (0, — )
1 3. Local maximum at (— j, |)
15. Saddle point at (0, 6, -3)
17. Local minimum at (0, 0, 0)
19. Saddle point at (-1, \, \)
21. (a) (0, -2) and (0,3)
(b) Local maximum at (0, —2);
local minimum at (0, 3)
23. (a) Minimum if a, b > 0; maximum if a, b < 0; saddle
point otherwise
(b) Minimum if a, b, c > 0; maximum if a, b, c < 0;
saddle point otherwise
(c) Minimum if a\, . . . , an > 0; maximum if a\, . . .,
an < 0; saddle point otherwise
25. Saddle points at (0,0), ^±-^,0^; local maxima at
(tt (~75' 7s)
27. Saddle points at (0, 0, 0, 0), (-V2, 2V2, 1 , -V2),
(V2, 2V2, -1, — n/2) {-42, -2V2, -1, 4%,
(a/2, -2V2, 1,a/2)
99 ^36 _48 _m
V 13 ' 13' 13,1
31 . 1 100 units of model X and 700 units of model Y
33. Maximum of 8 at (0, 0, 2);
minimum of — ^ at (— y , 3, |)
35. Maximum of 1 at (7r/2, 0), (jt/2, 2tt), (3tt/2, n);
minimum of — 1 at (3jt/2, 0), (3n/2, 2n), (jt/2, tt)
37. Maximum of 11 at (2, 0); minimum of | at (i, 1 )
39. Maximum of e6 at (0, 1, —2); minimum of eat all (x, y, z)
such that x2 + y2 - 2y + z2 + 4z = 0
41. (b) Maximum
43. (b) Neither
45. (b) Maximum
47. (a) Local maximum at (0, 0, 0)
(b) /(0, 0, 0) = e2 is a global maximum.
49. Global minimum of 2 + ln2 at (2,\). No global
maximum.
51. (a) {(1, y)\y eR}U{(*,2)|* eR)
(b) Maxima of 3 along critical points in (a).
53. (b) Critical points are (2, 1) and (0, —1).
588 Answers to Selected Exercises
(c)
z
Section 4.3
1 . (a) Minimize f(x, y) = x2 + y2 + (2x - 3y - A)2 to
find that the closest point is (=,— = , — =).
(b) Minimize f(x, y, z) = x2 + y2 + z2 subject to 2x —
3y-z=4.
3. (V2, 72) and (-^2, -72)
5. (1, |, 2), (3, 0,0), (0,2,0) and (0,0, 6)
' • V 1 1 ' n ' 11/
11- (±7C7r.^)"d(±^.-A.-^)
13. (a) (±yf,>),(0,±^)
(b) (=t-y/|> give maxima; ^0, ±^j) give minima.
17. (±1,0,0),(0,±1,0),(0,0,±1), (|,-|, j),
(_2 _2 _n
V 3' 3' 3/'
( 1 /H —1 _i /TT \ / i /TT _3 3 /TT\
^8V 2 ' 8' S\ 2 )'\ Sy 2 ' 8' 8V 2 J
19 / J_ _L krvg i-V2\ _j_ 1+yg 1+VI\
'7- v/2' 2 ' 2 VS' V2' 2 ' 2 J
21 . The numbers are 6, 6, 6.
23. Maximum value: 6. Minimum value: 0.
25. Height should be equal to diameter.
27. Locate at either (-2, 2, 1) or (-2, -2, 1).
29. Largest sphere has equation x2 + y2 + z2 = 2.
31. (|,2, f)
33. Highest point is (— 1, — 1, 2); lowest point is [j, i, |).
35. (a) (76, 76) and (-76, -76)
39. Nearest points are (75, — 75) and (— 75, 75);
farthest points are (5, 5) and (—5, —5).
41. (a) Critical point at (1,0)
(b) There is a minimum of 0 at (0, 0) and a maximum of
1 at (1,0).
y
(c) Vg = 0 at (0, 0)
43. (a) Hint:Checkthat3L/3/,- = c; - &(x)fon = \,...,k
and
dL_ = df_ _yj,dg[_
dxj dxj 4^ dxj
for j = 1, . . . , n.
Section 4.4
1 . 5x - ly + 14 = 0
3. (a) D(a, b) = E?=i(y« - (a/x, + b)f
(b) Minimize D with respect to a and b.
5. Hint: Let D(a, b, c) = IXiCv. - (ax2 + bxt + c))2 .
7. (b) There is a single, stable equilibrium point at
H.-i).
9. Single equilibrium point at (— 1 , |, |). There are no stable
equilibria.
Answers to Selected Exercises 589
1 1 . Produce 50,000 each of both the standard and executive
models and 100,000 deluxe models.
13. Irrigate only; Purchase 3333.3 gal of water.
1 5. (a) Invest $120,000 for capital equipment and $240,000
for labor.
(b) Hint: Note that L/K = 2.
True/False Exercises for Chapter 4
1 . True
3. True
5. True
7. False. (/ is most sensitive to changes in y.)
9. False
1 1 . True
1 3. False. (Consider the function f(x, y) = x2 + y2.)
15. True
1 7. False. (The point is not a critical point of the function.)
19. True
21 . False. (The critical point is a saddle point.)
23. False. (Extrema may also occur at points where g = c and
Vg = 0.)
25. False. (You will have to solve a system of 7 equations in
7 unknowns.)
27. True
29. False. (The equilibrium points are the critical points of the
potential function.)
Miscellaneous Exercises for Chapter 4
1 . r0 = 2h0
3. Price the Mocha at $2.70 per pound and the Kona at $5
per pound.
5. Maximum value of 4 at (1, — V3, 0). Minimum value of
-4at(-l, V3, 0).
7. (e)
(c) Maxima at Q=, ^,oV (-
j_
V2'
J= , 0^ ; minima
at (.7l<-71'0)' (-7I'7I'0); saddle P°mts at
(0,0, ±1)
11. l/(a2 + a2 + .-- + a2)
13. Dimensions are 4 (jc-direction) by 2\/2 (y-direction) by
2 (z-direction).
15. 1
17. a/3 by 6/3 by c/3
19. 3V5/8
21 . Hint: Minimize D2 = (x — xo)2 + (y — yo)2, where
(x, y) denotes a point on the line ax + by = d.
23. (a) Hint: Show that the maximum value occurs when
x2 = y2 = z2 = a2/3.
(b) Since f(x, y, z) = x2y2z2 is maximized when x2 =
y2 = z2 = a2/3, we must have x2y2z2 < (a2/3)3 =
((*2 + y2 + z2)/3)3.
(c) Hint: Since x\, Xz, ■ • ■ , x„ are assumed to be posi-
tive, we can write X; = yf for ; = 1 . Maximize
f(yi ,yi,---,yn) = y\y\ ■■■yl subject to y\ + y\ +
25. (a) A,i, A2
(a + c)± y/(a + c)2 - 4(ac - b2)
(b) Rewrite as k\ , ^2
Chapter 5
(a + c) ± V '(a - c)2 + 4fo2
9. (a)(0,O,±l),(il,0))(
j l rA
(■
V2'
Section 5.1
40
3
1.
3. 6(e - 1)
590 Answers to Selected Exercises
2e2 + e + 21n2
7. (a) Volume = f^(x2 + y2 + 2) dy dx = 26
9
11
(b) Volume = /Q2 f^(x2 + y2 + 2) dy
Jo I-ii^x2 + y4 sin^jc)dy d.v = 2 + 66/5^
The iterated integral gives the volume of the region
bounded by the graph of z = 16 — x2 — y2, the xy-plane,
and the planes x = 1, x = 3, y = —2, y = 2. The value
of the integral is .
13. The iterated integral gives the volume of the region
bounded by the graph of z = 4 — x2, the jcy-plane, and
the planes x = —2, x = 2, y = 0, y = 5. The value of the
integral is .
15. The iterated integral gives the volume of the region
bounded by the graph of z = 5 — \y\, the xy-plane, and
the planes x = —I, x = 2, y = —5, y = 5. The value of
the integral is 75.
Section 5.2
1. 0
3. (a) Check that f\fn " x3dydx
0.
(b) Hint: The region D is symmetric about the y-axis and
x3, is an odd function.
5. 4
2
(4,2)
1 -
— i — i — i — i —
12 3 4
7 1^2
' ■ 3
(-1,-1)'-!
9. 1 — COS 16 y
4
11. 3n/2 y
1-
Md.o) ,
J 2
-1-
13. 0
V = e*
2-
i
D
, (1, e)
1
-1
1 1
2 3 4
-1 -
\
y = -ex
4
3
99
20
128
5
15.
17.
19.
21.
23. Hint: Write / /s cf dA and ffRf dA as limits of appro-
priate Riemann sums.
25. Hint: Use the fact that |^"=i a< \ — \a'\-
27 i
29. Area is 7rafo.
31 1
6
18
33
35
37
39
50
3
11,664
16
3
Answers to Selected Exercises 591
41. (a) L f(x, y )dy = 2 regardless of whether x is rational
or irrational.
(b) Value is 2.
(c) 2
(d) Converges to 3.
(e) The Riemann sum has no uniquely determined limit.
(2,4)
(c) f*f£(2x + \)dxdy = 4
r4 r2-y/2
10 JO
y
y dx dy
16
3
4-"
(0,4)
3-
2-
V
= 4-2x
1-
1 —
\(2,0)
1
2
5. fifi(x + y)dydx=%
(3,9);
1- Jo £/2 eX dy dx + fi fl/2 e" dy dx = j + e>/2 - e
y
\x = l
1 -
x = y -
As — i
^ 1 1
1 2
9- C ' ' ydydx
16
3
(0,2)
= -M-y2/^
= V4 - y
/ 1-
(-2,0) (
\(2,0)
-2 -1
1 2
11 625
II. 12
1 ^ 896
io. 15
15. (1 - sinl)/2
17. (e27 - l)/6
19. (a) If your computer algebra system can provide a simple
answer, it should be 4(1 — cos 2).
(b) The integral with respect to y requires two applica-
tions of integration by parts.
(c) It should take the computer only a fraction of a second
to find that J*.1 f^y y2 cos (xy) dx dy = \ ( 1 — cos 2),
much faster than the evaluation in part (a).
21. (a) It is quite possible that the computer will be unable to
make the evaluation,
(b) With order reversed, the computer easily finds
J^fre—dydx = e-L
Section 5.4
1. 0
3. 1
5
7
1539
16
__5_
24
9. Volume :
11 12§
ii. 15
ra rVa2-.v2 r^Ja2-x2-y2
LaL^?I_^-^ dzdydx
13. 0
15 i
' 10
17. 81V3tt/8
592 Answers to Selected Exercises
19 15
21 ?9
3
23. V2tt/2
25. Jo Jo ~A /(*> y • z) rfz rfy
= Jo' fo~X fl% f(*> y. z)dydzdx
= fo fo~Z 11% f(x>y> z) dydxdz
= f-i fo~} If" /(*• y> z)dx dz dy
= fo f-%=i If' f(x,y,z)dxdy dz
z
X
27 • fo fy fo f(x' y ' z) dzdxdy
= ^ fo fz f(x> y. z) dydz dx
= ^ fz fz f(x,y,z) dydxdz
= f02 fo f2 f(x> y> z)dx dz dy
= fo2 f2 f2 f(x< y> z> dx dy dz
29. (a) Bottom surface is z = x2 + 3y2, top surface is z =
4 — y2; shadow in xy-plane is x2/4 + y2 < 1, y > 0.
00 fof2_i^j:i:^ + y')dzdxdy
(c) /0734//_^J(^3 + y3)^^^y
(d) fo If* ./^;:;<'S I Vs), A rfy dz +
(e) /_22 /,r2/4 /./^V + y3) ^ * «fa +
X?2 ft+xVi fo 4^(*3 + y3) Jy dzdx
Section 5.5
1 . (a) T(m, v) =
' 3
0 "
u
0
-1
V
(b) D = [0,3] x [-1,0]
3. D is the parallelogram with vertices (0, 0), (11, 2), (4, 3),
(15,5).
5. T takes W* onto the parallelepiped with vertices
(0, 0, 0), (3, 1, 5), (-1, -1, 3), (0, 2, -1), (2, 0, 8),
(3,3,4), (-1, 1,2), and (2,2, 7).
7. (a) D = {(x, y, z) | x2 + y2 + z2 < 1}
(b) D = {(x, y, z) | x2 + y2 + z2 < 1, x, y, z > 0}
(c) D = {(x, y, z) | 1 < x2 + y2 + z2 < 1, x, y, z > 0}
9. |02 /02 U5i;e"2 -\dudv = 8(e4 - l)/3
11. 7(e- 1/e)
13. 3tt
15. 486tt/5
17. 31n(^+ 1)
19. (16 - 3tt)/12
21. na2/4 if n is odd,
7ta2/2 if n is even.
23. 2 + JT/4
25. (jrsinl)/3
27. l(ln(V2+ 1) + V2- 1)
29. 7t(e4 - e + 1)
31. 48tt
33. ^[a3 -(a2 -fo2)3/2]
35. 2tt((1 - a2)eal + (b2 - \)ebl)
37. 8505tt/32
39. 4V2tt(5^-8)/3
41. 656tt/5
Section 5.6
1 . (a) 80 cases
(b) SI. 60
3. e2 - 2e + 1
5. (a) c, where c is the constant of proportionality,
(b) {(x,y,z) | x2 + y2 + z2 = 1}
30-3tt
30-5tt
9. 90 seconds
1 1 . If the plate is located at {(x, y) | x2 + y2 < a2, y > 0},
then(x,y) = (0, 4a/3jr).
13. (x,y)=(f,^)
1 5. (x, y) = ((7V3 + 8jt)/(3V3 + 4tt), 15/(Vl + An)\
17. (jc, 50 = (21/20,0)
19. (a) (*,y,z)=(i,0,§)
(b) (x,y,z)=(g,0, |)
21. (x,y,f)=(0,0,-l|)
23. 3o/8
25. (a) /v = /,, = If = ^
(b) r, = r„ = = 1/V5
27. (a) /- = 6561ir/4; r- = (3-v/3)/(2V2)
(b) /, = 8748^/35; r, = (373)777
Answers to Selected Exercises 593
29. 1496/135
31. 116tt/3
33. V = -3GMm(b2
Section 5.7
■a2)/(2(fc3-a3))
1. (a) J2,3 = -0.621375
b) Exact value is —0.619.
a) T2,i = 0.19825
b) Exact value is 0.197402.
a) 72,3 = 0.336123
b) Exact value is 0.33 1642.
7. (a) 52,2 = 0.00010125
b) Exact value is 0.0000972.
a) 52,2 = 0.331871
b) Exact value is 0.331642.
11. (a) S2,2 = 0.414325
b) Exact value is 0.414214.
. (a) io1 fo y/4x2 + 36y2 +ldydx
b) J4,4 = 3.52366
1 5. 52,2 is more accurate than T4 4.
17. (a) n must be at least 19.
b) n must be at least 1 .
19. (a) S2,i = -0.00390625
b) Exact value is — 1 /256.
c) The answers are the same because £2,2 = 0 by Theo-
rem 7.4.
21. 7-3,3 = 0.190978
23. 73,3 = 0.412888
25. 7-3,3 = 0.724061
True/False Exercises for Chapter 5
1 . False. (Not all rectangles must have sides parallel to the
coordinate axes.)
3. True
5. False
7. False
9. True
1 1 . True
1 3. False. (The value of the integral is 3.)
15. True
1 7. True. (The inner integral is zero because of symmetry.)
19. True
21 . False. (The integrals are opposites of one another.)
23. False. (A factor of r should appear in the integrand.)
25. False. (A factor of p is missing in the integrand.)
27. True
29. True
Miscellaneous Exercises for Chapter 5
1. 72;r
(a) f0 3 L%i%s fi/+y2 3 dz dy dx
-3 io J2r2+v2 ^ "z "y are two
and
possibilities,
(b) 8lV3 7r/4
-1 rVT^ p/9
dzdy dx
o» ,/;r ,c
1/3 foSC<P P1 sin(pdp dq> d6. Value is
16V2N
flit fim 1 1/3 p~i 2 ■ j j jr\ 1
J0 p smtp dp d(p d0 +
rliz rn/2
JO Jsi
2jt 9
7.
fo fo/4 io"SeCS rdr de dz
9. (sinl + sin2)/6
11. (a) r f^E^Ldydx
(b) E*
{(x, y)\x2 + y2
(c) Area = f^ fQl abr dr d6
13. Area is it.
15. -fln4
17. (In2)(tan-14
-)
4 I
y2 rV^
dw dz dy dx
19. ^f-Jl^f^
(b) 7t2a4/2
(c) Five-dimensional ball has volume 8jr2a5/15; six-
dimensional ball has volume jr3a6/6. The pattern is
not very clear from this information.
25. Within a disk of radius V5a/3 about the center, where a
is the radius of the hemisphere.
27. (a) 4(VT^7 - v^XvT3* - V8)
(b) 4
29. Integral does not converge.
31. -4ir/9
33. Converges when p > 1 and q > 1; value is
1
(p-\)(q-\)
35. Integral does not converge.
37. (a) Hint: Break up the integral into a sum of integrals
from 0 to 1 and from 1 to 00. Show convergence of
the improper integral from 1 to 00 by comparing it to
f^° e~x dx.
594 Answers to Selected Exercises
(b) Hint: Begin with JX.,e x y d A and use laws of ex-
ponents.
(c) 7r(l-e-fl2)
(d) 7T
(e) ^/n
39- (b) 4
41. C = a6/4
43. (a) C = ^
(b) |
45. 1 -e-'/t 0.2212
Chapter 6
Section 6.1
1. (a) 50 (b) 4
3. (35V35 - 17\/l7)/27
5. 5vT7
7. (53/2 + 87)/12
9. 2 11. | 13. (3e3 -5)/3
15. 4tt + 16;r2/3
17. 3tv 19. 1 - e"4;r
21. (a) /xF.ds=i,/yF.ds = -i
(b) y(r) = x(l — 2f). Thus, the images of x and y are the
same, although y traces the image in the opposite di-
rection to that of x. For these reasons, the results of
part (a) could have been anticipated.
23. 0
25. -M 27. 9 29. 0
31. -li
33. Hint: Use formula (3).
35. (a) 250(k/T3 ft-lb
(b) 7500 ft-lb
43. (a) 7.65625 (b) 7.5
Section 6.2
1 . §dDMdx + Ndy = f fD(Nx - My)dA = 8tt
3. <fSD M d.x + Ndy = f fD(Nx - My)dA = -4
5. fdD Mdx + N dy = f fD(Nx - My)dA = - Ujljz
7. (a) 0
(b) You will need to compute four separate line
integrals — one for each edge of the square.
9. -2
1 1 . Hint: Calculate \ fc —y dx + x dy, where C is the bound-
ary of R, oriented counterclockwise.
13. -45
15. (a) y
\ 1
-1 /
/ 1
CO TJ
1 9. Hint: Parametrize each line segment of the perimeter.
21. -12jt
23. (a) Hint: Show V • F = 0.
25. Directly apply Green's theorem.
27. The line integral is ±3 times the area of the rectangle,
where the sign depends on the orientation of the boundary.
29. Hint: uVu = (u dv/dx, u dv/dy). Then use Green's
theorem.
31. Hint: Begin with (b — ds and then use Green's
Jsd dn
theorem.
Section 6.3
1. (a) f
(b) ^
(c) No. Line integrals are not path-independent.
3. Not conservative
5. Not conservative
7. Conservative; f(x, y) = xe~y + cosjfy
9. Conservative; f(x, y) = 3x2y2 — jc3 + |y3
1 1 . Conservative; f(x, y, z) = 2x2yz^ — x2y + y2z
1 3. Conservative; f(x, y, z) = x2 + xy + sinyz
1 5. Conservative; f(x, y, z) = ex sin y + z3 + 2z
17. Not conservative
21 . N(x, y) = je2x + x3ey + u(y), where u is any function
of y of class C2.
23. N(x, y, z) = \x4 + z2 + h(y), h of class C1.
25. (a) Check that V x F = 0. Since F is of class C1 on all
of R3, by Theorem 3.5, F is conservative. A scalar
potential is f{x, y, z) = ^x3 + sin y sin z.
(b) | + sine sine2 — (sin l)2
27. 0
29. 6
31. -480
33. (a) Conservative on {(x, y) \ y > 0} and on
{(x, y)\y < 0}
Answers to Selected Exercises 595
(b) f(x,y) = (x2y2+x2 + l)/(2y2)
(c) 1
35. (a) F is conservative. (b) —2
37. Work =GMm(— — )
Vllxill ||x0||/
True/False Exercises for Chapter 6
1 . True
3. False. (It's negative.)
5. False. (The integral is 0.)
7. False. (There is equality only up to sign.)
9. True
1 1 . True
13. True
1 5. False. (The line integral could be ± fc ||F|| ds, depending
on whether F points in the same or the opposite direction
as C.)
1 7. False. (Let F = yi — xj and consider Green's theorem.)
1 9. False. (Under appropriate conditions, the integral is
f{B) - f(A).)
21. True
23 . False. (For the vector field to be conservative, the line inte-
gral must be zero for all closed curves, not just a particular
one.)
25. False. (The vector field (ex cos y sin z, ex sin y sin z,
ex cos y cos z) is not conservative.)
27. False. (The domain is not simply-connected.)
29. False. (/ is only defined up to at most a constant.)
Miscellaneous Exercises for Chapter 6
1. Break up C into n segments, each of length A.sj. By
continuity, / will be nearly constant on each segment,
so [/]avg ~ ELi f(ck)&Sk/ ELi Ask- (Here c* is any
point in the kth segment.) The formula follows after taking
limits as all As* — > 0.
3. la/n
5. 2
7. x = y = (*-*ri"-2f)a
y \ A(n-2sfl) )
9. (a) (b) 2
11 j _ 7769^17-1 _ / 7769Vf7-T
1 ~~ 840 ' r* - V 1190^17-70
13. /- =27711, rz =
15. (a) Xf g(f{e) cos 6, f(6) sinfl V(/(0))2 + {f'(6)f d6
(b) yi0[(e1811 - l)/9+ 12(1 -e12ir)/37]
17. 6tt
19. K = 2n
21 . | — cos 1 — sin 1
23. Hint: Use the formula Area = j J3D —y dx + x dy.
25. Hint: x = ffDx d A/area of D,
y = ffDy dA /area of D. Now apply Green's theorem to
<fdD x2 dy and §aD xy dy.
27. x = y = -±4
29. Hint: Use the result of Exercise 28 twice.
35. (a) Both the line integral and the double integral are zero,
(b) No. The double integral is not defined properly over
the disk because F is undefined at the origin.
W 7/fl[^(^)-i(^)]^ = o = /cF-^ +
fc F • ds. Since fc F • ds = — 2n, the result follows.
37. (a) 0 "
(b) Apply the divergence theorem.
39. /xF- ds = |X-VV • ds = -V(B)+ V(A). Now use
Exercise 38.
Chapter 7
Section 7.1
1. (a) -i-4j + 2k
(b) x + 4y - 2z = 5
3. y + 4z = 4
y
(b) Yes
(c) 4x - 2y - z = 3
7. (a) Smooth except at (0, 0,0). Tangent plane at (1,V3, 4)
has equation 2x + 2^/3y — z = 4.
(b) S is a paraboloid.
(c) z = x2 + y2
(d) Yes it does. At (0, 0, 0), the tangent plane has equation
z = 0.
9. Hint: Consider x2 + y2.
1 1 . (a)-(c) All versions give the equation
-x + y + \/2z = 1.
13. X(s, t) = (2cos s, r, 2 sins), 0 < s < 2jt, -1 < t < 3
15. Xi(s, t) = (s, t, Vs2 + t2+ lY
X2(s, f) = ^, f, -V.?2 + f2 + l), (i, G R2
17. (a) y2z = t2s2=x2
596 Answers to Selected Exercises
x
(d) Points on the positive z-axis
(e) -Ax + Sy + z = 4
(f) (0, 0, 1) = X(±l, 0); tangent planes have equations
x ± y = 0.
19. (a) At the point (a, g(a, c), c), the tangent plane has equa-
tion gx(a, c)(x -a)-(y- g(a, c)) + gz(a, c)(z -
c) = 0.
(b) At the point (h(b, c), b, c), the tangent plane has equa-
tion -(x - h(b, c)) + hy(b, c)(y - b) + hz(b, c){z -
c) = 0.
21. {x e R3 | x = (1, 0, \) + (s - 1)(1,2, 0) +
0 + 1)(0, 1, -2)} or x=s,y = 2s + t-\,z = -It -
1. To verify consistency with Exercise 5(b), check that
the points given by the parametric equations all lie in the
plane determined by the equation in Exercise 5(b).
23. V6tt
25. 4tWo2 - b2
27. (653^2 - 173/2)tt/24
29. yTTa -areaofD
31. 16a2
Section 7.2
1. 26V3/3
3. 1
r 640
O. 3
7. (a) 4jra4 (b) 47za4/3
9. (a) Parametrize the cylinder as x = 2cosf, y = 2sinf,
z = s with —2 < s < 2, 0 < / < lit . The integral
evaluates to — 647r.
(b) The integral is —4 • (surface area of S) = —64n.
11. 0
13. 297tt/2
15. 36tt
17. 0
19. 27rfl3/3
21. —Tta2
(b) ni = (x/(V2z), y/(V2z), -1/V2), n2 = (x, y, 0)
(c) -1456jt/3
29. (a)
z
(b) (a cosi, a sins, 0)
(c) N(s, 0) = (a cos s (2 cos(s/2) + sin(s /2)),
a sins (2 cos(s/2) + 8^(^/2)),
a(2 sin(s/2) — cos(.?/2))).
From this, we see that N(0, 0) = (2a, 0, —a), while
N(2tt, 0) = (-2a, 0, a). Therefore, the Klein bottle
cannot be orientable, since the normal vector along
the s-coordinate curve at t = 0 changes direction.
Section 7.3
1.0 3. 0
5. 0 7. 0
9. 4jr(b2 - a2)
11. 45tt
13. (a) Hint: Use the double angle formula.
(b) -f
15. 1/2
17. 625tt/8
25. (a) na
(b) it a
(c) The answers in parts (a) and (b) are the same; the
three flat quarter-circles that are part of 3D do not
contribute anything to jjsVf-ndS.
Answers to Selected Exercises 597
29. (a) §sF-dS
« Fz(r, 9,z + Az/2)r A9 Ar
- Fz(r, 6,z- Az/2)r AO Ar
+ Fr(r + Ar/2, 9, z)(r + Ar/2)A9 Az
- Fr(r - Ar/2, 9, z)(r - Ar/2)A9 Az
+ F9(r, 9 + A6/2, z)Ar Az
- Fe(r, 9 - A6/2, z)Ar Az
Section 7.4
1 . Hint: Use Gauss's theorem and the product rule for
V-(/Vg).
3. (a) Hint: Use Green's first formula with f = g.
(b) Hint: Use part (a) and the fact that V/ • V/ =
IIV/H2.
7 . Hint: Use the argument in Exercise 6 and the product rule
forV-(A;Vr).
9. (a) Hint: Use Gauss's theorem and Exercise 8.
(b) Heat flows into D from the inner sphere and out
through the outer sphere at the same rate.
1 1 . Hint: Use Ampere's and Gauss's laws.
15. (b) In each case k = [jlq€q.
1 9. Hint: Apply Gauss's theorem to P = E x B, then Fara-
day's and Ampere's laws.
21. Hint: Apply the arguments used in Exercise 20 to each
component integral of B.
(a) (Vz2+ 1 cos (Wz2 + 1 sin6», z)
(b) (a*Js2 + 1 cos?, b\l s1 + 1 sin?, cs)
(a) (a sirup cos9, bsincp sinO, c coscp)
(bU2Vo*
b2c2 sin4 <p cos2
+a2b2 cos2 cp sin2 (p
+ a2c2 sin4 cp sin2 (
dtp dG
(a) (s cos t, f(s), s sinf)
(b) t
n
10
11.
13. (a/2, a/2, a/2)
15. (0,0, a/3)
17. (a) 15>/57r/2
(b) V572
(c) Iz = 62V5jrk/5, rz = 793/35
19. (a) Ix = Iy = 2jrab8(3a2 + 2b2)/3
(b) r, = ry = V(3fl2 + 2b2)/6
23. Hint: Use Stokes's theorem.
25. Use Exercise 24.
31 . Hint: Show V • (x/||x||3) = 0 where defined.
35. V • F 0, so there is no vector potential.
39. Hint: Calculate V x (E + dA/dt).
Chapter 8
True/False Exercises for Chapter 7
1 . True
3. True. (Let u = s3 and v = tanf.)
5. False. (The limits of integration are not correct.)
7. False. (The value of the integral is 24.)
9. True
1 1 . False. (The integral has value 32jt.)
1 3. False. (The value is 0.)
1 5. False. (The surface must be connected.)
17. True. (The result follows from Stokes's theorem.)
19. True. (Use Gauss's theorem.)
21 . False. (Gauss's theorem implies that the integral is at most
twice the surface area.)
23. True
25. False. (Should be the flux of the curl of F.)
27. True. (Apply Green's first formula.)
29. False. (/ is determined up to addition of a harmonic
function.)
Miscellaneous Exercises for Chapter 7
1.(a) C (b) E (c) A
(d) D (e) F (f) B
Section 8.1
1.-2 3. 6 5. 182
7. -370
9. —6a\b^ + 6a^b\ + 02^4 — 04^2
11. 14cosx - 7sinz + lly2 + 33
13. 6 (ev cos y + (y2 + 2)e2z)
1 5. 2xy dx A dy A dz
17. 3^3(^:3 — X4) dx\ A dx2 A dx-$ A dx4
1 9. (x\eXAXs — X1JC2X3 cos X5) dx\ A dxi A dx^ A dx$ A dx^
Section 8.2
1. Hint: Show linear independence of T9l, Tg2, Tg3 by
solving the vector equation CiT^, + C2T#2 + c^TS] = 0
for c\, C2, C3.
3. X: [0, 2tt) x [0, 2jt) x [0, 2jt) x [1, 2] x [1,2] R6,
X(6»i, 62, 6*3, l2, h) = (xu y\, x2, V2, xi, y3), where x\ =
3 cos 9\, yi=3sin#i, X2 = 3 cos#i + h cos 6*2, y2 =
3 sin 61 + 12 sin 62, X3 = 3 cos#i + h cos 02 + h cos03,
y3 = 3 sin 9\ + 1% sin #2 + ^3 sin #3
7. -2jt
9. 4jt2
11. (a) Sx(u)(T„1,T„2,T„3) = m1 > 0 for 0 < u\ < JJ
which is where the parametrization is smooth.
598 Answers to Selected Exercises
(b) Parametrize dM in two pieces as
Yi:[0,V5] x [0, 2tt) R3,
Yi(ii, S2) = (si cosi2, *i sinj2, $\ — 6) and
(c) V
Y2:[0, V5] x [0, lit) R3,
*2) = (*1 COS 52,
(2«i cos S2 , 2s 1 sin ^2,
Y2CS1, 52) = Csi COS52, *i sinj2, 4 — ij).
1)
V2
lAs\ + 1
(2*i cos 52, 25i sin 52, 1)
4j? + 1
13. 32?r
15 ^
u. 2
Section 8.3
1 . dco = exyz(yz dx + xzdy + xy dz)
3. dco = —ydx A dy
5. dco = (x + 2yz)<ix A dy A dz
7. d&> = 2(X] — X2 + X3 — • ■ • + (— l)"+1x„) rfxi A JX2 A
• • • A dx„
9. F(x, z) = xz + Cz + D\, G(x,y) = xy + Cy + D2,
where C, D\, D2 are arbitrary constants.
True/False Exercises for Chapter 8
1 . True 3 . True
5. True 7. True
9. True
11. False. (X(l, 1, -1) = X(l, 1, 1), so X is not one-one
on D.)
1 3. False. (The agreement is only up to sign.)
1 5. False. (This is only true if n is even.)
17. True
19. True. {dco would be an (n + l)-form, and there are no
nonzero ones on R".)
Miscellaneous Exercises for Chapter 8
5. 0
7. (a) V x (V/) = 0
(b) V • (V x F) = 0
9. Hint: Consider fM d(w Arj).
Index
n, xv
U, xv
V (del operator), 227-228
C, xv
C, XV
A
acceleration, 190
normal component of, 2 1 5
tangential component of, 215
accumulation point, 103
addition of vectors, see vector(s),
addition, 2
adjoint matrix, 61
Ampere's law, 516-517, 518
static case, 516
time-varying case, 517, 518
angle between vectors, 2 1 , 49
angular momentum, 243
angular speed, 34
angular velocity vector, 221
antiderivative, xvii
Archimedes' principle, 506
arclength function, 205, 217
arclength parameter, 205-207
area
zero, 319
area element
Cartesian, 364, 370
general, 370
polar, 364, 370
area of a surface, 464^167
arithmetic mean, 308
arithmetic-geometric inequality, 308
astroid, 219
average value, 373-377
of a function on a surface, 525
B
ball
closed, 101
open, 101
basis, see vector(s), standard basis, 9
Bernoulli, Johann, 194
Bezier curve, 239
biharmonic function, 187
binormal spherical image, 242
binormal vector, see vector(s), binormal,
212
BMI, 155
body mass index, 155
boundary
of a manifold, 541
boundary of a set, 102
Brahe, Tycho, 193
budget hyperplane, 56
buoyant force, 506
c
Cauchy-Riemann equations, 242
Cauchy-Schwarz inequality, 50, 293
Cavalieri's principle, 310
center of mass
continuous
in R, 380
in R2, 380
in R3, 382
continuous, 379-383
discrete, 377-379
of a wire, 451
centroid, 383
chain rule, 142-153
in one variable, 143-144
in several variables, 144-153
change of variables
in double integrals, 357, 370
in triple integrals, 365, 371
charge density, 514
charge distribution, 514
C°°, 138
circular helix, 190
circulation, 413, 476
circulation density, 498
C* 138
Clairaut, Alexis, 137
class C°°, 138
class C*, 138
closed box, 337
closed rectangle, 310, 314
closed set, 102
clothoid, 241
Cobb-Douglas production function, 302
codomain of a function, 82
cofactor, 57
column vector, 52
commodity bundle, 56
compact set, 271
component functions, 85
conduction current density, 517
conductivity, 522
connected, 441
conservative vector field, 297, 441
constraint, 279
constraint equation, 278
continuity
piecewise, 318
continuity equation, 516, 520
for current densities, 520
in fluid dynamics, 520
contour curves, 87
definition, 88
coordinate axes
in R2, xv
in R3, xv
coordinate curves
of a manifold, 538
coordinate transformations, 349-355
coordinates
Cartesian, 62
conversions
between Cartesian and cylindrical,
65
between Cartesian and polar, 64
between Cartesian and spherical,
68
between cylindrical and spherical,
68
cylindrical, 65
hyperspherical, 72-73
on R2, xv
on R3, xv
on the real line, xv
polar, 62
rectangular, see coordinates,
Cartesian, 62
spherical, 66
Copernicus, Nicholas, 193
critical point, 182, 265
constrained, 279
degenerate, 268, 288
cross product, 27-36
and determinant, 3 1
applications, 32-35
definition, 28
inR", 61
properties, 29
curl, 227,229-231,497
in cylindrical coordinates, 233, 508
in spherical coordinates, 235, 509
current density, 514
conduction, 517
displacement, 517
curvature, 209-210, 216, 217
total, 452
radius of, 240
curve, 419
convex, 452
Bezier, 239
closed, 419
simple, 419
cusp, 80
cycloid
curtate, 17
prolate, 18
D
Darboux formulas, 221
Darboux rotation vector, 22 1
definite integral, see integral, definite,
xvi
degree of a polynomial, 141
600 Index
degree of a term, 141
del operator (V), 227-228
derivative, xv, 124, 127-128
directional, 160, 161
maximization of, 163
minimization of, 163
of a vector field, 236
exterior, see exterior derivative, 553
linearity of, 134
normal, see normal derivative, 453,
507, 520
partial derivative
mixed, 136
of higher order, 136-138
partial derivative of f with respect to
xt, 117
determinant, 30, 56
properties, 60
differentiability
and continuity, 122, 127
definition, 121
in general, 126
of a function of three or more
variables, 123-127
of a function of two variables,
120-123
differential geometry, 202
differential operators, 152
differentiating under the integral sign,
405
direction cosines, 26
directional derivative, see derivative,
directional, 160
displacement current density, 517
displacement vector, 5
distance
between parallel planes, 46
between point and line, 45^16, 77
between point and plane, 77
between skew lines, 46-47
divergence, 228-229, 496
in cylindrical coordinates, 233,
507
in spherical coordinates, 235, 508
divergence theorem, see Gauss's
theorem, 493
in the plane, 432
domain of a function, 82
dot product, 18-25, 36
definition, 19
properties, 19
double integral, 316
in polar coordinates, 360-362
E
eigenvalue, 309
eigenvector, 309
electric field
of a continuous charge distribution,
514
of a single point charge, 512
electromotive force, 517
elementary region
in space, 340-343
definition, 340
in the plane, 321-327
definition, 321
ellipsoid, 94
parametrized, 468, 525
elliptic cone, 94
elliptic paraboloid, 94
endowment vector, 56
energy
kinetic, 297
potential, 297
epicycles, 193
epicycloid, 80, 437
epitrochoid, 80
equation of continuity, see continuity
equation, 520
equation of first variation, 227
equilibrium point, 298
stable, 298
unstable, 298
equipotential line, 224
equipotential set, 224
equipotential surface, 224
Euler's formula, 188
evolute, 240
extension (/ext), 322, 343
exterior derivative, 553
of a £-form, 553
of a 0-form, 553
exterior product, 534-535
properties of, 535
extrema
absolute, 264
constrained local, 288
global, 264, 270-274
local, 263
second derivative test, 268
extreme value theorem, 27 1
F
Faraday's law, 517-518
Fenchel's theorem, 452
field
scalar, see scalar field, 222
vector, see vector field, 221
first variation
equation of, see equation of first
variation, 227
flow line, 225
flow of a vector field, 227
flux, 432, 476
flux density, 498
force
buoyant, 506
Frenet-Serret formulas, 217-219
Fresnel integrals, 241
frustum, 526
Fubini's theorem, 314, 319, 339
functions, 82-94
average value, 374
along a curve, 45 1
codomain, 82
component, see component functions,
85
continuous, 109
algebraic properties, 1 1 1
domain, 82
extension of (/ext), 322, 343
homogeneous, 187, 309
injective, 83
linear, 55
mean value of, 374
of more than one variable, 84-94
graphing, 87-94
quadric surfaces, see quadric
surfaces, 93
one-one, see functions, injective, 83
onto, see functions, surjective, 83
partial, 1 16
polynomial, 106
potential, 223, 441
range, 83
scalar-valued, 84
smooth, 138
surjective, 83
fundamental theorem of calculus,
xvii
G
gauge freedom, 528
Gauss's law, 512-514, 518
differential form, 514
integral form, 514
Gauss's theorem, 433, 493^196
implied by generalized Stokes's
theorem, 558-559
proof of, 503-505
generalized Stokes's theorem, see
Stokes's theorem, generalized,
553
geometric mean, 308
global extrema, 264, 270-274
on compact regions, 270-274
gradient, 124, 158, 227-228
in cylindrical coordinates, 233
in spherical coordinates, 235
gradient field, 223
gravitational potential, 388
Green's first formula, 510
Green's first identity, 453
Green's formulas, 510
Green's second formula, 510
Green's second identity, 453
Green's theorem, 429
implied by generalized Stokes's
theorem, 557
proof of, 433-436
vector reformulation of, 43 1
Green's third formula, 510
gyration, radius of, 385, 451, 526
Index 601
H
harmonic function, 142, 243, 453, 511,
520
head-to-tail addition, 4
heat equation, 142, 520, 521
uniqueness of solutions to, 521
heat flux density, 489, 520
helicoid, 142, 468
helix, 85, 190
Hessian, 255
Hessian criterion
for constrained extrema, 286-290
for extrema, 265-270
Hilbert matrix, 79
homogeneous function, 187, 309
hyperbolic paraboloid, 90, 94
hyperboloid
parametrized, 524
hyperboloid of one sheet, 94
hyperboloid of two sheets, 94-95
hyperplane
budget, 56
hypersphere, 73, 167, 293, 308
hyperspherical coordinates, 72-73
hypersurface, 73, 124
hypocycloid, 80
hypotrochoid, 80
ideal gas law, 96, 158
identity matrix, 59, 177
implicit function theorem, 168-170
general case, 170
improper integral, 405^406
incremental change, 249
injectivity, 83
inner product, see dot product, 19
inR",49
integrability, 316, 339
integral
definite, xvi
double, 316
improper, 405
in polar coordinates, 360-362
improper, 405-406
iterated, 3 1 1
line
scalar, see line integral, scalar, 409
vector, see line integral, vector, 411
of a differential form, 536-542
of a A>form over a fc-manifold, 541
of a 2-form, 537
properties of
linearity, 320
monotonicity, 320
triple, 338
improper, 405, 522
in cylindrical coordinates, 366-368
in spherical coordinates, 368-370
integration by parts for differential
forms, 562
intersection of sets, xv
inverse function theorem, 172
inverse matrix, 61
involute, 15, 240
isobars, 224
isolated point, 103
isoquants, 302
isotherms, 224
iterated integral, 311
J
Jacobi identity, 39, 78
Jacobian, 172, 356, 365
judo, 6-7
K
A'-form
basic, 533
general, 533
Kepler, Johannes, 193
kinetic energy, 297
Klein bottle, 489
L
Lagrange multipliers, 280-290
lamina, 380
Laplace's equation, 141, 453
Laplacian operator, 157, 187, 236
inversion formula for, 511
law of conservation of energy, 454
least squares approximation,
293-297
Leibniz's rule, 405
length, 204
level curves, 87
definition, 88
limit(s), 97-111
algebraic properties, 106
geometric interpretation, 103-104
intuitive definition, 98
rigorous definition, 99-100
uniqueness, 106
line
parametric equations of,
see parametric equations, of a
line, 13
line integral
differential form, 415
numerical approximation of, 421^126
path-independent, 439, 440
scalar, 409
vector, 411
line segment, 76
linear combination, 543
linear independence, 284, 539
linear mapping, see mapping, linear, 55
linear regression, 293
linear span, 543
lines
skew, 46
Lorentz force, 428
M
magnetic field
of a moving charge distribution, 515
of a moving point charge, 514
magnetic monopoles, 516, 518
manifold, 202, 538
underlying, 538
boundary of a, 541
general, 551
smooth, 539
mapping, linear, 55
matrices, 30-32, 51-58
adjoint, 61
cofactor, 57
cofactor expansion, 57
determinant, 56
elementary row operations, 60
Hilbert, 79
identity, 59, 177
inverse, 61
invertible, 61, 177
matrix product, 53
properties, 54
minor, 57
nilpotent, 79
nonsingular, 61
scalar multiplication, 53
properties, 53
symmetric, 267, 309
transpose, 54
triangular, 60
matrix of partial derivatives, 125
maximum, see extrema, 263
Maxwell's equations, 512-518
mean value theorem, 329
for double integrals, 399, 526
for triple integrals, 497
general version, 260
mean value theorem for integrals, 260
method of least squares, 293
minimal surface, 142
minimum, see extrema, 263
minor, 57
Mobius strip, 480
moment, 377
first, 377, 525
of inertia, 384-386, 451, 526
second, 384-386
total, 377
monopoles
magnetic, see magnetic monopoles,
516
moving frame, 211,217
multiple regression, 297
N
n -dimensional volume, 560
negative definite, 267, 309
neighborhood, 102, 122
nephroid, 81
net force, 8
602 Index
Newton's method, 176-181
Newton, Isaac, 194
nilpotent matrix, 79
nonorientable, 479
normal derivative, 453, 507, 520
normal line
to a plane curve, 175
to a surface in space, 175
normal plane, see plane, normal, 22 1
normal spherical image, 242
normalization, 23
numerical integration
for functions of one variable, 388-391
for functions of two variables,
391^100
Monte Carlo method, 399
o
octant, xv
1-form
basic, 530
general, 530
one-one, see injectivity, 83
one-sided, see nonorientable, 479
onto, see surjectivity, 83
open set, 1 02
opposite path, 417
orientable, 479, 543
orientation, 420, 479, 543
compatible, 543
consistent, 490, 546
induced, 490, 546
oriented, 420, 543
origin
in R2, xv
in R3, xv
orthogonal, 21
osculating plane, see plane, osculating,
211,220
Ostrogradsky's theorem, see Green's
theorem, 436
Ostrogradsky, Mikhail, 436
P
parallel axis theorem, 407
parallelogram, 7
parallelogram law, 4
parametric equations
of a line
symmetric form, 12
parametric equations, 10
of a cycloid, 14-15
of an involute, 15
of a line, 9-13
in terms of a point and a direction,
10
in terms of two points, 12
parametrization, 419
parametrized surfaces, 455-467
definition, 455
smooth, 460
area of, 464^167
piecewise smooth, 463
partial derivative, 117
mixed, 136
of higher order, 136-138
partial functions, 116
partition, xvi, 314, 338
path, 189
closed, 419
endpoints, 189
flow line, 225
intrinsic quantities, 217
nonintrinsic quantities, 216
nonrectifiable, 205
opposite, 417
rectifiable, 205
simple, 419
tangent line, 191
velocity vector, 190
path independence, 439, 440
permeability of free space, 514
permittivity of free space, 512
perpendicular, see orthogonal, 2 1
perpendicular bisector, 76
plane
coordinate equation for, 40-43
normal, 22 1
osculating, 211, 220
parametric equations for, 43^15
rectifying, 221
planetary motion, Kepler's laws of,
193-200
polynomial, 106, 107
position vector, 3
positive definite, 267, 309
potential, 441
vector, see vector potential, 528
potential energy, 297
potential function, see functions,
potential, 223
potential theory, 142
Poynting vector field, 522
principal minors, sequence of, 268
principal normal vector, see vector(s),
principal normal, 211
probability density, 406
joint, 406
product rule
for scalar- valued functions, 135
nonexistence of a general form, 135
of a scalar-valued and a vector-valued
function, 136
projection of a vector, 21-24
Pythagorean theorem, 59
Q
quadrant, xv
quadratic form, 267, 309
negative definite, 267, 309
positive definite, 267, 309
quadric surfaces, 93-95
ellipsoid, 94
elliptic cone, 94
elliptic paraboloid, 94
hyperbolic paraboloid, 94
hyperboloid of one sheet, 94
hyperboloid of two sheets, 94-95
quotient rule
for scalar-valued functions, 135
nonexistence of a general form, 135
R
radius of curvature, 240
radius of gyration, 385, 451, 526
random variables
independent, 407
range of a function, 83
rectifying plane, see plane, rectifying,
221
reparametrization, 416
of a /c-manifold, 549
of a manifold, 543
of a surface, 477
orientation-preserving, 416, 418, 478,
549
orientation-reversing, 416, 418, 478,
549
reparametrization of a path, 205-207
resultant force, 8
Riemann sum, xvi, 315, 338
right-hand rule, 28, 36
right-handed system, xv
row vector, 52
s
s -coordinate curve, 458
saddle point, 265
scalar field, 222
scalar line integral, see line integral,
scalar, 409
scalar multiplication, 2, 4, 36
inR2 andR3,2
inR",49
properties, 3
scalar potentials, 445-447
scalar product, see dot product, 19
scalars, 1
Scherk's surface, 142
second derivative test
for constrained extrema, 286-290
for extrema, 265-270
section of a surface, 88
set
boundary of a, 102
closed, 102
compact, 271
open, 102
simply-connected region, 442
Simpson's rule
for functions of one variable, 390
for functions of two variables, 394
smooth function, 138
Index 603
Snell's law of refraction, 308
solid angle, 527
span, 543
linear, 543
speed, 190
sphere, 92
spiral of Cornu, 241
Spirograph, 79
standard basis vectors, see vector(s),
standard basis, 9
standard normal vector, 460
steradians, 527
stereographic projection, 524
Stokes's theorem, 431, 490^193
generalized, 553
and the fundamental theorem of
calculus, 560
implied by generalized Stokes's
theorem, 557-558
proof of, 500-503
strake, 241
subset, xv
surface
minimal, 142
of revolution, see surface of
revolution, 525
oriented, 481
surface area, 464-467
surface area element, 485
surface integrals, 469^488
scalar, 470, 485
vector, 474, 486
surface of revolution, 525
surjectivity, 83
symmetric matrix, 267, 309
T
r-coordinate curve, 458
tangent hyperplane, 124, 166
tangent line, to a path, 191
tangent plane, 118, 164-168
equation, 166
parametric equations for, 469
to a smooth parametrized surface, 462
tangent spherical image, 242
Taylor polynomial, 244, 246
first-order, 248
higher-order, 256
second-order, 254
Taylor's theorem
in one variable, 244-247
in several variables
first-order formula, 247-248
formulas for polynomials of order
greater than two, 256
second-order formula, 252-254
Lagrange's form of the remainder, 257
telegrapher's equation, 522
thermal conductivity, 520
tofu, 263, 340
topology, 101
torque, 34, 243, 385
torsion, 212-214, 217-219
torus, 202, 459
total differential, 249
tractrix, 240
transpose, 54
trapezoidal rule
for functions of one variable, 388
for functions of two variables, 392
triangle inequality, 5 1
triple integral, 338
in cylindrical coordinates, 366-368
in spherical coordinates, 368-370
2-form
basic, 531
general, 532
two-sided, see orientable, 479
u
underlying manifold, 538
underlying surface, 455
union of sets, xv
unit vector, 22, 23
utility, 301
utility function, 1 86
V
vector field, 189, 221
conservative, 297, 441
curl of, 229-231
divergence of, 228-229
gradient, 223, 297, 441
incompressible, 229
irrotational, 23 1
radially symmetric, 453
solenoidal, 229
vector line integral, see line integral,
vector, 411
vector potential, 516, 528
vector product, see cross product, 28
vector projection, 21-24
vector surface integral element, 486
vector(s), 1
addition, 4
inR",49
properties, 2
algebraic notion, 1-3
angle between, 21
inR",49
binormal, 212, 217
cross product
properties, 29
cross product, 27-36
and determinant, 3 1
applications, 32-35
definition, 28
definition
inR2 andR3, 1
inR",49
difference, 4
displacement, see displacement
vector, 5
distance, 49
dot product, 18-25, 36
definition, 19
properties, 19
equality
inR2 andR3, 1
inR", 49
geometric notion, 3-7
gradient, 158
inner product
inR", 49
length, 19, 49
magnitude, see vector(s), length, 19
norm, see vector(s), length, 19
notation, 1
position vector, see position vector, 3
principal normal, 211,217
projection of, 21-24
scalar multiplication, 2, 4, 36
inR2 andR3, 2
inR", 49
properties, 3
standard basis, 9
for cylindrical coordinates, 71-72
for spherical coordinates, 71-72
inR", 50
unit, 22, 23
unit normal, outward, 432
unit tangent, 207, 412
zero, 2
vectors
standard normal, 460
velocity, 190
volume, 310
zero, 339
volume element
Cartesian, 371
general, 365, 371
in cylindrical coordinates, 367, 371
in spherical coordinates, 369, 371
w
wave equation, 187
wedge product, see exterior product, 534
Whitney umbrella, 468
windchill, 84, 185
work, 26, 411
Z
zero area, 319
zero vector, 2
zero volume, 339
0-form, 530
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